Solve for θ sin(theta)^2=cos(theta)^2+1

$\sin^{2} (θ) = \cos^{2} (θ) + 1$

Move all the expressions to the left side of the equation.

Subtract $\cos^{2} (θ)$ from both sides of the equation.

$\sin^{2} (θ) - \cos^{2} (θ) = 1$

Subtract $1$ from both sides of the equation.

$\sin^{2} (θ) - \cos^{2} (θ) - 1 = 0$

Simplify $\sin^{2} (θ) - \cos^{2} (θ) - 1$ .

Move $- 1$ .

$\sin^{2} (θ) - 1 - \cos^{2} (θ) = 0$

Reorder $\sin^{2} (θ)$ and $- 1$ .

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

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

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

Factor $- 1$ out of $\sin^{2} (θ)$ .

$- 1 (1) - 1 (- \sin^{2} (θ)) - \cos^{2} (θ) = 0$

Factor $- 1$ out of $- 1 (1) - 1 (- \sin^{2} (θ))$ .

$- 1 (1 - \sin^{2} (θ)) - \cos^{2} (θ) = 0$

Rewrite $- 1 (1 - \sin^{2} (θ))$ as $- (1 - \sin^{2} (θ))$ .

$- (1 - \sin^{2} (θ)) - \cos^{2} (θ) = 0$

Apply pythagorean identity.

$- \cos^{2} (θ) - \cos^{2} (θ) = 0$

Subtract $\cos^{2} (θ)$ from $- \cos^{2} (θ)$ .

$- 2 \cos^{2} (θ) = 0$

Solve for $θ$ .

Divide each term in $- 2 \cos^{2} (θ) = 0$ by $- 2$ and simplify.

Divide each term in $- 2 \cos^{2} (θ) = 0$ by $- 2$ .

$\frac{- 2 \cos^{2} (θ)}{- 2} = \frac{0}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 \cos^{2} (θ)}{- 2} = \frac{0}{- 2}$

Divide $\cos^{2} (θ)$ by $1$ .

$\cos^{2} (θ) = \frac{0}{- 2}$

Simplify the right side.

Divide $0$ by $- 2$ .

$\cos^{2} (θ) = 0$

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

$\cos (θ) = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$\cos (θ) = \pm \sqrt{0^{2}}$

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

$\cos (θ) = \pm 0$

Plus or minus $0$ is $0$ .

$\cos (θ) = 0$

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

$θ = \arccos (0)$

Simplify the right side.

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$θ = \frac{π}{2}$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

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

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

Find the period of $\cos (θ)$ .

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 $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Consolidate the answers.

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

Solve for θ sin(theta)^2=cos(theta)^2+1

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