Calculus Examples

$\sin (3 x) = \cos (2 x)$

Step 1

Subtract $\cos (2 x)$ from both sides of the equation.

$\sin (3 x) - \cos (2 x) = 0$

Step 2

Simplify each term.

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Step 2.1

Apply the sine triple-angle identity.

$- 4 \sin^{3} (x) + 3 \sin (x) - \cos (2 x) = 0$

Step 2.2

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

$- 4 \sin^{3} (x) + 3 \sin (x) - (1 - 2 \sin^{2} (x)) = 0$

Step 2.3

Apply the distributive property.

$- 4 \sin^{3} (x) + 3 \sin (x) - 1 \cdot 1 - (- 2 \sin^{2} (x)) = 0$

Step 2.4

Multiply $- 1$ by $1$ .

$- 4 \sin^{3} (x) + 3 \sin (x) - 1 - (- 2 \sin^{2} (x)) = 0$

Step 2.5

Multiply $- 2$ by $- 1$ .

$- 4 \sin^{3} (x) + 3 \sin (x) - 1 + 2 \sin^{2} (x) = 0$

Step 3

Factor $- 4 \sin^{3} (x) + 3 \sin (x) - 1 + 2 \sin^{2} (x)$ .

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Step 3.1

Reorder terms.

$- 4 \sin^{3} (x) + 2 \sin^{2} (x) + 3 \sin (x) - 1 = 0$

Step 3.2

Factor $- 4 \sin^{3} (x) + 2 \sin^{2} (x) + 3 \sin (x) - 1$ using the rational roots test.

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Step 3.2.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1$

$q = \pm 1, \pm 4, \pm 2$

Step 3.2.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.25, \pm 0.5$

Step 3.2.3

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

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Step 3.2.3.1

Substitute $1$ into the polynomial.

$- 4 \cdot 1^{3} + 2 \cdot 1^{2} + 3 \cdot 1 - 1$

Step 3.2.3.2

Raise $1$ to the power of $3$ .

$- 4 \cdot 1 + 2 \cdot 1^{2} + 3 \cdot 1 - 1$

Step 3.2.3.3

Multiply $- 4$ by $1$ .

$- 4 + 2 \cdot 1^{2} + 3 \cdot 1 - 1$

Step 3.2.3.4

Raise $1$ to the power of $2$ .

$- 4 + 2 \cdot 1 + 3 \cdot 1 - 1$

Step 3.2.3.5

Multiply $2$ by $1$ .

$- 4 + 2 + 3 \cdot 1 - 1$

Step 3.2.3.6

Add $- 4$ and $2$ .

$- 2 + 3 \cdot 1 - 1$

Step 3.2.3.7

Multiply $3$ by $1$ .

$- 2 + 3 - 1$

Step 3.2.3.8

Add $- 2$ and $3$ .

$1 - 1$

Step 3.2.3.9

Subtract $1$ from $1$ .

$0$

Step 3.2.4

Since $1$ is a known root, divide the polynomial by $\sin (x) - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- 4 \sin^{3} (x) + 2 \sin^{2} (x) + 3 \sin (x) - 1}{\sin (x) - 1}$

Step 3.2.5

Divide $- 4 \sin^{3} (x) + 2 \sin^{2} (x) + 3 \sin (x) - 1$ by $\sin (x) - 1$ .

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Step 3.2.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$\sin (x)$

$1$

$4 \sin^{3} (x)$

$2 \sin^{2} (x)$

$3 \sin (x)$

$1$

Step 3.2.5.2

Divide the highest order term in the dividend $- 4 \sin^{3} (x)$ by the highest order term in divisor $\sin (x)$ .

				-	$4 \sin^{2} (x)$
	$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$

Step 3.2.5.3

Multiply the new quotient term by the divisor.

			-	$4 \sin^{2} (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			-	$4 \sin^{3} (x)$	+	$4 \sin^{2} (x)$

Step 3.2.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 \sin^{3} (x) + 4 \sin^{2} (x)$

			-	$4 \sin^{2} (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$

Step 3.2.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$4 \sin^{2} (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$

Step 3.2.5.6

Pull the next terms from the original dividend down into the current dividend.

			-	$4 \sin^{2} (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$

Step 3.2.5.7

Divide the highest order term in the dividend $- 2 \sin^{2} (x)$ by the highest order term in divisor $\sin (x)$ .

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$

Step 3.2.5.8

Multiply the new quotient term by the divisor.

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$
					-	$2 \sin^{2} (x)$	+	$2 \sin (x)$

Step 3.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 \sin^{2} (x) + 2 \sin (x)$

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$
					+	$2 \sin^{2} (x)$	-	$2 \sin (x)$

Step 3.2.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$
					+	$2 \sin^{2} (x)$	-	$2 \sin (x)$
							+	$\sin (x)$

Step 3.2.5.11

Pull the next terms from the original dividend down into the current dividend.

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$
					+	$2 \sin^{2} (x)$	-	$2 \sin (x)$
							+	$\sin (x)$	-	$1$

Step 3.2.5.12

Divide the highest order term in the dividend $\sin (x)$ by the highest order term in divisor $\sin (x)$ .

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$	+	$1$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$
					+	$2 \sin^{2} (x)$	-	$2 \sin (x)$
							+	$\sin (x)$	-	$1$

Step 3.2.5.13

Multiply the new quotient term by the divisor.

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$	+	$1$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$
					+	$2 \sin^{2} (x)$	-	$2 \sin (x)$
							+	$\sin (x)$	-	$1$
							+	$\sin (x)$	-	$1$

Step 3.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $\sin (x) - 1$

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$	+	$1$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$
					+	$2 \sin^{2} (x)$	-	$2 \sin (x)$
							+	$\sin (x)$	-	$1$
							-	$\sin (x)$	+	$1$

Step 3.2.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$4 \sin^{2} (x)$	-	$2 \sin (x)$	+	$1$
$\sin (x)$	-	$1$	-	$4 \sin^{3} (x)$	+	$2 \sin^{2} (x)$	+	$3 \sin (x)$	-	$1$
			+	$4 \sin^{3} (x)$	-	$4 \sin^{2} (x)$
					-	$2 \sin^{2} (x)$	+	$3 \sin (x)$
					+	$2 \sin^{2} (x)$	-	$2 \sin (x)$
							+	$\sin (x)$	-	$1$
							-	$\sin (x)$	+	$1$
										$0$

Step 3.2.5.16

Since the remander is $0$ , the final answer is the quotient.

$- 4 \sin^{2} (x) - 2 \sin (x) + 1$

Step 3.2.6

Write $- 4 \sin^{3} (x) + 2 \sin^{2} (x) + 3 \sin (x) - 1$ as a set of factors.

$(\sin (x) - 1) (- 4 \sin^{2} (x) - 2 \sin (x) + 1) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sin (x) - 1 = 0$

$- 4 \sin^{2} (x) - 2 \sin (x) + 1 = 0$

Step 5

Set $\sin (x) - 1$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $\sin (x) - 1$ equal to $0$ .

$\sin (x) - 1 = 0$

Step 5.2

Solve $\sin (x) - 1 = 0$ for $x$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$\sin (x) = 1$

Step 5.2.2

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (1)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

The exact value of $\arcsin (1)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 5.2.4

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{2}$

Step 5.2.5

Simplify $π - \frac{π}{2}$ .

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Step 5.2.5.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = π \cdot \frac{2}{2} - \frac{π}{2}$

Step 5.2.5.2

Combine fractions.

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Step 5.2.5.2.1

Combine $π$ and $\frac{2}{2}$ .

$x = \frac{π \cdot 2}{2} - \frac{π}{2}$

Step 5.2.5.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 5.2.5.3

Simplify the numerator.

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Step 5.2.5.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 5.2.5.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 5.2.6

Find the period of $\sin (x)$ .

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Step 5.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 5.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 5.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 5.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.7

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{2} + 2 π n$ , for any integer $n$

Step 6

Set $- 4 \sin^{2} (x) - 2 \sin (x) + 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $- 4 \sin^{2} (x) - 2 \sin (x) + 1$ equal to $0$ .

$- 4 \sin^{2} (x) - 2 \sin (x) + 1 = 0$

Step 6.2

Solve $- 4 \sin^{2} (x) - 2 \sin (x) + 1 = 0$ for $x$ .

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Step 6.2.1

Substitute $u$ for $\sin (x)$ .

$- 4 {(u)}^{2} - 2 u + 1 = 0$

Step 6.2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6.2.3

Substitute the values $a = - 4$ , $b = - 2$ , and $c = 1$ into the quadratic formula and solve for $u$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (- 4 \cdot 1)}}{2 \cdot - 4}$

Step 6.2.4

Simplify.

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Step 6.2.4.1

Simplify the numerator.

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Step 6.2.4.1.1

Raise $- 2$ to the power of $2$ .

$u = \frac{2 \pm \sqrt{4 - 4 \cdot - 4 \cdot 1}}{2 \cdot - 4}$

Step 6.2.4.1.2

Multiply $- 4 \cdot - 4 \cdot 1$ .

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Step 6.2.4.1.2.1

Multiply $- 4$ by $- 4$ .

$u = \frac{2 \pm \sqrt{4 + 16 \cdot 1}}{2 \cdot - 4}$

Step 6.2.4.1.2.2

Multiply $16$ by $1$ .

$u = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot - 4}$

Step 6.2.4.1.3

Add $4$ and $16$ .

$u = \frac{2 \pm \sqrt{20}}{2 \cdot - 4}$

Step 6.2.4.1.4

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 6.2.4.1.4.1

Factor $4$ out of $20$ .

$u = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot - 4}$

Step 6.2.4.1.4.2

Rewrite $4$ as $2^{2}$ .

$u = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot - 4}$

Step 6.2.4.1.5

Pull terms out from under the radical.

$u = \frac{2 \pm 2 \sqrt{5}}{2 \cdot - 4}$

Step 6.2.4.2

Multiply $2$ by $- 4$ .

$u = \frac{2 \pm 2 \sqrt{5}}{- 8}$

Step 6.2.4.3

Simplify $\frac{2 \pm 2 \sqrt{5}}{- 8}$ .

$u = \frac{1 \pm \sqrt{5}}{- 4}$

Step 6.2.4.4

Move the negative in front of the fraction.

$u = - \frac{1 \pm \sqrt{5}}{4}$

Step 6.2.5

The final answer is the combination of both solutions.

$u = - \frac{1 + \sqrt{5}}{4}, - \frac{1 - \sqrt{5}}{4}$

Step 6.2.6

Substitute $\sin (x)$ for $u$ .

$\sin (x) = - \frac{1 + \sqrt{5}}{4}, - \frac{1 - \sqrt{5}}{4}$

Step 6.2.7

Set up each of the solutions to solve for $x$ .

$\sin (x) = - \frac{1 + \sqrt{5}}{4}$

$\sin (x) = - \frac{1 - \sqrt{5}}{4}$

Step 6.2.8

Solve for $x$ in $\sin (x) = - \frac{1 + \sqrt{5}}{4}$ .

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Step 6.2.8.1

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (- \frac{1 + \sqrt{5}}{4})$

Step 6.2.8.2

Simplify the right side.

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Step 6.2.8.2.1

Evaluate $\arcsin (- \frac{1 + \sqrt{5}}{4})$ .

$x = - \frac{3 π}{10}$

Step 6.2.8.3

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{3 π}{10} + π$

Step 6.2.8.4

Simplify the expression to find the second solution.

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Step 6.2.8.4.1

Subtract $2 π$ from $2 π + \frac{3 π}{10} + π$ .

$x = 2 π + \frac{3 π}{10} + π - 2 π$

Step 6.2.8.4.2

The resulting angle of $\frac{13 π}{10}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{3 π}{10} + π$ .

$x = \frac{13 π}{10}$

Step 6.2.8.5

Find the period of $\sin (x)$ .

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Step 6.2.8.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 6.2.8.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 6.2.8.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 6.2.8.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.8.6

Add $2 π$ to every negative angle to get positive angles.

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Step 6.2.8.6.1

Add $2 π$ to $- \frac{3 π}{10}$ to find the positive angle.

$- \frac{3 π}{10} + 2 π$

Step 6.2.8.6.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$2 π \cdot \frac{10}{10} - \frac{3 π}{10}$

Step 6.2.8.6.3

Combine fractions.

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Step 6.2.8.6.3.1

Combine $2 π$ and $\frac{10}{10}$ .

$\frac{2 π \cdot 10}{10} - \frac{3 π}{10}$

Step 6.2.8.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 10 - 3 π}{10}$

Step 6.2.8.6.4

Simplify the numerator.

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Step 6.2.8.6.4.1

Multiply $10$ by $2$ .

$\frac{20 π - 3 π}{10}$

Step 6.2.8.6.4.2

Subtract $3 π$ from $20 π$ .

$\frac{17 π}{10}$

Step 6.2.8.6.5

List the new angles.

$x = \frac{17 π}{10}$

Step 6.2.8.7

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{13 π}{10} + 2 π n, \frac{17 π}{10} + 2 π n$ , for any integer $n$

Step 6.2.9

Solve for $x$ in $\sin (x) = - \frac{1 - \sqrt{5}}{4}$ .

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Step 6.2.9.1

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (- \frac{1 - \sqrt{5}}{4})$

Step 6.2.9.2

Simplify the right side.

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Step 6.2.9.2.1

Evaluate $\arcsin (- \frac{1 - \sqrt{5}}{4})$ .

$x = \frac{π}{10}$

Step 6.2.9.3

$x = 2 π - \frac{π}{10} + π$

Step 6.2.9.4

Simplify the expression to find the second solution.

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Step 6.2.9.4.1

Subtract $2 π$ from $2 π - \frac{π}{10} + π$ .

$x = 2 π - \frac{π}{10} + π - 2 π$

Step 6.2.9.4.2

The resulting angle of $\frac{9 π}{10}$ is positive, less than $2 π$ , and coterminal with $2 π - \frac{π}{10} + π$ .

$x = \frac{9 π}{10}$

Step 6.2.9.5

Find the period of $\sin (x)$ .

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Step 6.2.9.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 6.2.9.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 6.2.9.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 6.2.9.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.9.6

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{10} + 2 π n, \frac{9 π}{10} + 2 π n$ , for any integer $n$

Step 6.2.10

List all of the solutions.

$x = \frac{13 π}{10} + 2 π n, \frac{17 π}{10} + 2 π n, \frac{π}{10} + 2 π n, \frac{9 π}{10} + 2 π n$ , for any integer $n$

Step 7

The final solution is all the values that make $(\sin (x) - 1) (- 4 \sin^{2} (x) - 2 \sin (x) + 1) = 0$ true.

$x = \frac{π}{2} + 2 π n, \frac{13 π}{10} + 2 π n, \frac{17 π}{10} + 2 π n, \frac{π}{10} + 2 π n, \frac{9 π}{10} + 2 π n$ , for any integer $n$

Step 8

Consolidate the answers.

$x = \frac{π}{10} + \frac{2 π n}{5}$ , for any integer $n$