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

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Trigonometry

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

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

$2 \sin (x) \cos (x) - \cos (x) = 0$

Factor $\cos (x)$ out of $2 \sin (x) \cos (x) - \cos (x)$ .

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Factor $\cos (x)$ out of $2 \sin (x) \cos (x)$ .

$\cos (x) (2 \sin (x)) - \cos (x) = 0$

Factor $\cos (x)$ out of $- \cos (x)$ .

$\cos (x) (2 \sin (x)) + \cos (x) \cdot - 1 = 0$

Factor $\cos (x)$ out of $\cos (x) (2 \sin (x)) + \cos (x) \cdot - 1$ .

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cos (x) = 0$

$2 \sin (x) - 1 = 0$

Set $\cos (x)$ equal to $0$ and solve for $x$ .

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Set $\cos (x)$ equal to $0$ .

$\cos (x) = 0$

Solve $\cos (x) = 0$ for $x$ .

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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 $n$

Set $2 \sin (x) - 1$ equal to $0$ and solve for $x$ .

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Set $2 \sin (x) - 1$ equal to $0$ .

$2 \sin (x) - 1 = 0$

Solve $2 \sin (x) - 1 = 0$ for $x$ .

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Add $1$ to both sides of the equation.

$2 \sin (x) = 1$

Divide each term in $2 \sin (x) = 1$ by $2$ and simplify.

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Divide each term in $2 \sin (x) = 1$ by $2$ .

$\frac{2 \sin (x)}{2} = \frac{1}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \sin (x)}{2} = \frac{1}{2}$

Divide $\sin (x)$ by $1$ .

$\sin (x) = \frac{1}{2}$

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

$x = \arcsin (\frac{1}{2})$

Simplify the right side.

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The exact value of $\arcsin (\frac{1}{2})$ is $30$ .

$x = 30$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$x = 180 - 30$

Subtract $30$ from $180$ .

$x = 150$

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

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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 $\sin (x)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 30 + 360 n, 150 + 360 n$ , for any integer $n$

The final solution is all the values that make $\cos (x) (2 \sin (x) - 1) = 0$ true.

$x = 90 + 360 n, 270 + 360 n, 30 + 360 n, 150 + 360 n$ , for any integer $n$

Consolidate $90 + 360 n$ and $270 + 360 n$ to $90 + 180 n$ .

$x = 90 + 180 n, 30 + 360 n, 150 + 360 n$ , for any integer $n$

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