Trigonometry Examples

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

Step 1

Substitute $u$ for $\sin (2 x)$ .

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

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

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

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

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

Step 4.1.2

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

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 4.1.2.2

Multiply $- 4$ by $1$ .

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

Step 4.1.3

Subtract $4$ from $9$ .

$u = \frac{3 \pm \sqrt{5}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$u = \frac{3 \pm \sqrt{5}}{2}$

Step 5

The final answer is the combination of both solutions.

$u = \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}$

Step 6

Substitute $\sin (2 x)$ for $u$ .

$\sin (2 x) = \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}$

Step 7

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

$\sin (2 x) = \frac{3 + \sqrt{5}}{2}$

$\sin (2 x) = \frac{3 - \sqrt{5}}{2}$

Step 8

Solve for $x$ in $\sin (2 x) = \frac{3 + \sqrt{5}}{2}$ .

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Step 8.1

The range of sine is $- 1 \leq y \leq 1$ . Since $\frac{3 + \sqrt{5}}{2}$ does not fall in this range, there is no solution.

No solution

Step 9

Solve for $x$ in $\sin (2 x) = \frac{3 - \sqrt{5}}{2}$ .

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Step 9.1

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

$2 x = \arcsin (\frac{3 - \sqrt{5}}{2})$

Step 9.2

Simplify the right side.

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Step 9.2.1

Evaluate $\arcsin (\frac{3 - \sqrt{5}}{2})$ .

$2 x = 0.39192267$

Step 9.3

Divide each term in $2 x = 0.39192267$ by $2$ and simplify.

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Step 9.3.1

Divide each term in $2 x = 0.39192267$ by $2$ .

$\frac{2 x}{2} = \frac{0.39192267}{2}$

Step 9.3.2

Simplify the left side.

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Step 9.3.2.1

Cancel the common factor of $2$ .

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Step 9.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0.39192267}{2}$

Step 9.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0.39192267}{2}$

Step 9.3.3

Simplify the right side.

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Step 9.3.3.1

Divide $0.39192267$ by $2$ .

$x = 0.19596133$

Step 9.4

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.

$2 x = (3.14159265) - 0.39192267$

Step 9.5

Solve for $x$ .

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Step 9.5.1

Subtract $0.39192267$ from $3.14159265$ .

$2 x = 2.74966997$

Step 9.5.2

Divide each term in $2 x = 2.74966997$ by $2$ and simplify.

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Step 9.5.2.1

Divide each term in $2 x = 2.74966997$ by $2$ .

$\frac{2 x}{2} = \frac{2.74966997}{2}$

Step 9.5.2.2

Simplify the left side.

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Step 9.5.2.2.1

Cancel the common factor of $2$ .

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Step 9.5.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{2.74966997}{2}$

Step 9.5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2.74966997}{2}$

Step 9.5.2.3

Simplify the right side.

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Step 9.5.2.3.1

Divide $2.74966997$ by $2$ .

$x = 1.37483498$

Step 9.6

Find the period of $\sin (2 x)$ .

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Step 9.6.1

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

$\frac{2 π}{| b |}$

Step 9.6.2

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 9.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Step 9.6.4

Cancel the common factor of $2$ .

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Step 9.6.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 9.6.4.2

Divide $π$ by $1$ .

$π$

Step 9.7

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

$x = 0.19596133 + π n, 1.37483498 + π n$ , for any integer $n$

Step 10

List all of the solutions.

$x = 0.19596133 + π n, 1.37483498 + π n$ , for any integer $n$