Solve for x in Degrees 2sin(x)^2-cos(x)^2=2

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Trigonometry

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

Subtract $2$ from both sides of the equation.

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

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

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Move $- 2$ .

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

Reorder $2 \sin^{2} (x)$ and $- 2$ .

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

Factor $- 2$ out of $- 2$ .

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

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

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

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

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

Apply pythagorean identity.

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

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

$- 3 \cos^{2} (x) = 0$

Solve for $x$ .

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Divide each term in $- 3 \cos^{2} (x) = 0$ by $- 3$ and simplify.

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Divide each term in $- 3 \cos^{2} (x) = 0$ by $- 3$ .

$\frac{- 3 \cos^{2} (x)}{- 3} = \frac{0}{- 3}$

Simplify the left side.

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Cancel the common factor of $- 3$ .

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Cancel the common factor.

$\frac{- 3 \cos^{2} (x)}{- 3} = \frac{0}{- 3}$

Divide $\cos^{2} (x)$ by $1$ .

$\cos^{2} (x) = \frac{0}{- 3}$

Simplify the right side.

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Divide $0$ by $- 3$ .

$\cos^{2} (x) = 0$

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

$\cos (x) = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

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

$\cos (x) = \pm \sqrt{0^{2}}$

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

$\cos (x) = \pm 0$

Plus or minus $0$ is $0$ .

$\cos (x) = 0$

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

$x = \arccos (0)$

Simplify the right side.

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The exact value of $\arccos (0)$ is $90$ .

$x = 90$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$x = 360 - 90$

Subtract $90$ from $360$ .

$x = 270$

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

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

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

$\frac{360}{| 1 |}$

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

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\cos (x)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 90 + 360 n, 270 + 360 n$ , for any integer $n$

Consolidate the answers.

$x = 90 + 180 n$ , for any integer $n$

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