Solve for θ cos(2theta)+10sin(theta)^2=7

$\cos (2 θ) + 10 \sin^{2} (θ) = 7$

Step 1

Subtract $7$ from both sides of the equation.

$\cos (2 θ) + 10 \sin^{2} (θ) - 7 = 0$

Step 2

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

$1 - 2 \sin^{2} (θ) + 10 \sin^{2} (θ) - 7 = 0$

Step 3

Move all terms not containing $θ$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$- 2 \sin^{2} (θ) + 10 \sin^{2} (θ) - 7 = - 1$

Add $7$ to both sides of the equation.

$- 2 \sin^{2} (θ) + 10 \sin^{2} (θ) = - 1 + 7$

Step 4

Simplify the left side.

Add $- 2 \sin^{2} (θ)$ and $10 \sin^{2} (θ)$ .

$8 \sin^{2} (θ) = - 1 + 7$

Step 5

Simplify the right side.

Add $- 1$ and $7$ .

$8 \sin^{2} (θ) = 6$

Step 6

Solve the equation for $θ$ .

Divide each term in $8 \sin^{2} (θ) = 6$ by $8$ and simplify.

Divide each term in $8 \sin^{2} (θ) = 6$ by $8$ .

$\frac{8 \sin^{2} (θ)}{8} = \frac{6}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 \sin^{2} (θ)}{8} = \frac{6}{8}$

Divide $\sin^{2} (θ)$ by $1$ .

$\sin^{2} (θ) = \frac{6}{8}$

Simplify the right side.

Cancel the common factor of $6$ and $8$ .

Factor $2$ out of $6$ .

$\sin^{2} (θ) = \frac{2 (3)}{8}$

Cancel the common factors.

Factor $2$ out of $8$ .

$\sin^{2} (θ) = \frac{2 \cdot 3}{2 \cdot 4}$

Cancel the common factor.

$\sin^{2} (θ) = \frac{2 \cdot 3}{2 \cdot 4}$

Rewrite the expression.

$\sin^{2} (θ) = \frac{3}{4}$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$\sin (θ) = \pm \sqrt{\frac{3}{4}}$

Simplify $\pm \sqrt{\frac{3}{4}}$ .

Rewrite $\sqrt{\frac{3}{4}}$ as $\frac{\sqrt{3}}{\sqrt{4}}$ .

$\sin (θ) = \pm \frac{\sqrt{3}}{\sqrt{4}}$

Simplify the denominator.

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

$\sin (θ) = \pm \frac{\sqrt{3}}{\sqrt{2^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$\sin (θ) = \pm \frac{\sqrt{3}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$\sin (θ) = \frac{\sqrt{3}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$\sin (θ) = - \frac{\sqrt{3}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$\sin (θ) = \frac{\sqrt{3}}{2}, - \frac{\sqrt{3}}{2}$

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

$\sin (θ) = \frac{\sqrt{3}}{2}$

$\sin (θ) = - \frac{\sqrt{3}}{2}$

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

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

$θ = \arcsin (\frac{\sqrt{3}}{2})$

Simplify the right side.

The exact value of $\arcsin (\frac{\sqrt{3}}{2})$ is $\frac{π}{3}$ .

$θ = \frac{π}{3}$

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.

$θ = π - \frac{π}{3}$

Simplify $π - \frac{π}{3}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

Combine $π$ and $\frac{3}{3}$ .

$θ = \frac{π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 3 - π}{3}$

Simplify the numerator.

Move $3$ to the left of $π$ .

$θ = \frac{3 \cdot π - π}{3}$

Subtract $π$ from $3 π$ .

$θ = \frac{2 π}{3}$

Find the period of $\sin (θ)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

$θ = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n$ , for any integer $n$

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

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

$θ = \arcsin (- \frac{\sqrt{3}}{2})$

Simplify the right side.

The exact value of $\arcsin (- \frac{\sqrt{3}}{2})$ is $- \frac{π}{3}$ .

$θ = - \frac{π}{3}$

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.

$θ = 2 π + \frac{π}{3} + π$

Simplify the expression to find the second solution.

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

$θ = 2 π + \frac{π}{3} + π - 2 π$

The resulting angle of $\frac{4 π}{3}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{3} + π$ .

$θ = \frac{4 π}{3}$

Find the period of $\sin (θ)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

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

$- \frac{π}{3} + 2 π$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$\frac{2 π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$\frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$\frac{5 π}{3}$

List the new angles.

$θ = \frac{5 π}{3}$

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

$θ = \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

List all of the solutions.

$θ = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Consolidate the solutions.

Consolidate $\frac{π}{3} + 2 π n$ and $\frac{4 π}{3} + 2 π n$ to $\frac{π}{3} + π n$ .

$θ = \frac{π}{3} + π n, \frac{2 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Consolidate $\frac{2 π}{3} + 2 π n$ and $\frac{5 π}{3} + 2 π n$ to $\frac{2 π}{3} + π n$ .

$θ = \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer $n$

Solve for θ cos(2theta)+10sin(theta)^2=7

Related Questions

Support LCMGCF.com by disabling your adblocker.