Trigonometry Examples

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

Step 1

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

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

Step 2

Subtract $1$ from both sides of the equation.

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

Step 3

Subtract $3 \sin (x)$ from both sides of the equation.

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

Step 4

Solve the equation for $x$ .

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Step 4.1

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

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

Step 4.2

Add $1$ to both sides of the equation.

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

Step 4.3

Use the quadratic formula to find the solutions.

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

Step 4.4

Substitute the values $a = - 2$ , $b = - 3$ , and $c = 1$ into the quadratic formula and solve for $u$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (- 2 \cdot 1)}}{2 \cdot - 2}$

Step 4.5

Simplify.

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Step 4.5.1

Simplify the numerator.

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Step 4.5.1.1

Raise $- 3$ to the power of $2$ .

$u = \frac{3 \pm \sqrt{9 - 4 \cdot - 2 \cdot 1}}{2 \cdot - 2}$

Step 4.5.1.2

Multiply $- 4 \cdot - 2 \cdot 1$ .

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Step 4.5.1.2.1

Multiply $- 4$ by $- 2$ .

$u = \frac{3 \pm \sqrt{9 + 8 \cdot 1}}{2 \cdot - 2}$

Step 4.5.1.2.2

Multiply $8$ by $1$ .

$u = \frac{3 \pm \sqrt{9 + 8}}{2 \cdot - 2}$

Step 4.5.1.3

Add $9$ and $8$ .

$u = \frac{3 \pm \sqrt{17}}{2 \cdot - 2}$

Step 4.5.2

Multiply $2$ by $- 2$ .

$u = \frac{3 \pm \sqrt{17}}{- 4}$

Step 4.5.3

Move the negative in front of the fraction.

$u = - \frac{3 \pm \sqrt{17}}{4}$

Step 4.6

The final answer is the combination of both solutions.

$u = - \frac{3 + \sqrt{17}}{4}, - \frac{3 - \sqrt{17}}{4}$

Step 4.7

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

$\sin (x) = - \frac{3 + \sqrt{17}}{4}, - \frac{3 - \sqrt{17}}{4}$

Step 4.8

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

$\sin (x) = - \frac{3 + \sqrt{17}}{4}$

$\sin (x) = - \frac{3 - \sqrt{17}}{4}$

Step 4.9

Solve for $x$ in $\sin (x) = - \frac{3 + \sqrt{17}}{4}$ .

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Step 4.9.1

The range of sine is $- 1 \leq y \leq 1$ . Since $- \frac{3 + \sqrt{17}}{4}$ does not fall in this range, there is no solution.

No solution

Step 4.10

Solve for $x$ in $\sin (x) = - \frac{3 - \sqrt{17}}{4}$ .

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Step 4.10.1

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

$x = \arcsin (- \frac{3 - \sqrt{17}}{4})$

Step 4.10.2

Simplify the right side.

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Step 4.10.2.1

Evaluate $\arcsin (- \frac{3 - \sqrt{17}}{4})$ .

$x = 0.28460296$

Step 4.10.3

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.

$x = 2 (3.14159265) - 0.28460296 + 3.14159265$

Step 4.10.4

Simplify the expression to find the second solution.

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Step 4.10.4.1

Subtract $2 π$ from $2 (3.14159265) - 0.28460296 + 3.14159265$ .

$x = 2 (3.14159265) - 0.28460296 + 3.14159265 - 2 π$

Step 4.10.4.2

The resulting angle of $2.85698969$ is positive, less than $2 π$ , and coterminal with $2 (3.14159265) - 0.28460296 + 3.14159265$ .

$x = 2.85698969$

Step 4.10.5

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

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Step 4.10.5.1

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

$\frac{2 π}{| b |}$

Step 4.10.5.2

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

$\frac{2 π}{| 1 |}$

Step 4.10.5.3

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

$\frac{2 π}{1}$

Step 4.10.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.10.6

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

$x = 0.28460296 + 2 π n, 2.85698969 + 2 π n$ , for any integer $n$

Step 4.11

List all of the solutions.

$x = 0.28460296 + 2 π n, 2.85698969 + 2 π n$ , for any integer $n$