Trigonometry Examples

$\cos^{2} (x) = 3 (1 - \sin (x))$

Step 1

Replace the $\cos^{2} (x)$ with $1 - \sin^{2} (x)$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

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

Step 2

Reorder the polynomial.

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

Step 3

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

$- {(u)}^{2} + 1 = 3 (1 - (u))$

Step 4

Simplify $3 (1 - u)$ .

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Step 4.1

Rewrite.

$- u^{2} + 1 = 0 + 0 + 3 (1 - u)$

Step 4.2

Simplify by adding zeros.

$- u^{2} + 1 = 3 (1 - u)$

Step 4.3

Apply the distributive property.

$- u^{2} + 1 = 3 \cdot 1 + 3 (- u)$

Step 4.4

Multiply.

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Step 4.4.1

Multiply $3$ by $1$ .

$- u^{2} + 1 = 3 + 3 (- u)$

Step 4.4.2

Multiply $- 1$ by $3$ .

$- u^{2} + 1 = 3 - 3 u$

Step 5

Add $3 u$ to both sides of the equation.

$- u^{2} + 1 + 3 u = 3$

Step 6

Subtract $3$ from both sides of the equation.

$- u^{2} + 1 + 3 u - 3 = 0$

Step 7

Subtract $3$ from $1$ .

$- u^{2} + 3 u - 2 = 0$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $- 1$ out of $- u^{2} + 3 u - 2$ .

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Step 8.1.1

Factor $- 1$ out of $- u^{2}$ .

$- u^{2} + 3 u - 2 = 0$

Step 8.1.2

Factor $- 1$ out of $3 u$ .

$- u^{2} - (- 3 u) - 2 = 0$

Step 8.1.3

Rewrite $- 2$ as $- 1 (2)$ .

$- u^{2} - (- 3 u) - 1 \cdot 2 = 0$

Step 8.1.4

Factor $- 1$ out of $- u^{2} - (- 3 u)$ .

$- (u^{2} - 3 u) - 1 \cdot 2 = 0$

Step 8.1.5

Factor $- 1$ out of $- (u^{2} - 3 u) - 1 (2)$ .

$- (u^{2} - 3 u + 2) = 0$

Step 8.2

Factor.

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Step 8.2.1

Factor $u^{2} - 3 u + 2$ using the AC method.

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Step 8.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $- 3$ .

$- 2, - 1$

Step 8.2.1.2

Write the factored form using these integers.

$- ((u - 2) (u - 1)) = 0$

Step 8.2.2

Remove unnecessary parentheses.

$- (u - 2) (u - 1) = 0$

Step 9

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

$u - 2 = 0$

$u - 1 = 0$

Step 10

Set $u - 2$ equal to $0$ and solve for $u$ .

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Step 10.1

Set $u - 2$ equal to $0$ .

$u - 2 = 0$

Step 10.2

Add $2$ to both sides of the equation.

$u = 2$

Step 11

Set $u - 1$ equal to $0$ and solve for $u$ .

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Step 11.1

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 11.2

Add $1$ to both sides of the equation.

$u = 1$

Step 12

The final solution is all the values that make $- (u - 2) (u - 1) = 0$ true.

$u = 2, 1$

Step 13

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

$\sin (x) = 2, 1$

Step 14

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

$\sin (x) = 2$

$\sin (x) = 1$

Step 15

Solve for $x$ in $\sin (x) = 2$ .

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Step 15.1

The range of sine is $- 1 \leq y \leq 1$ . Since $2$ does not fall in this range, there is no solution.

No solution

Step 16

Solve for $x$ in $\sin (x) = 1$ .

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Step 16.1

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

$x = \arcsin (1)$

Step 16.2

Simplify the right side.

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Step 16.2.1

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

$x = \frac{π}{2}$

Step 16.3

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 16.4

Simplify $π - \frac{π}{2}$ .

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Step 16.4.1

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 16.4.2

Combine fractions.

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Step 16.4.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 16.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 16.4.3

Simplify the numerator.

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Step 16.4.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 16.4.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 16.5

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

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Step 16.5.1

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

$\frac{2 π}{| b |}$

Step 16.5.2

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

$\frac{2 π}{| 1 |}$

Step 16.5.3

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

$\frac{2 π}{1}$

Step 16.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 16.6

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

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

Step 17

List all of the solutions.

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