Solve for x 2sin(x)cos(x) = square root of 2cos(x)

$2 \sin (x) \cos (x) = \sqrt{2} \cos (x)$

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

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

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

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

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

Factor $\cos (x)$ out of $- \sqrt{2} \cos (x)$ .

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

Factor $\cos (x)$ out of $\cos (x) (2 \sin (x)) + \cos (x) (- \sqrt{2})$ .

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

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

$\cos (x) = 0$

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

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

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

$\cos (x) = 0$

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

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

$x = \arccos (0)$

Simplify the right side.

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

$x = \frac{π}{2}$

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.

$x = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

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

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

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

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

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

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

Set $2 \sin (x) - \sqrt{2}$ equal to $0$ .

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

Solve $2 \sin (x) - \sqrt{2} = 0$ for $x$ .

Add $\sqrt{2}$ to both sides of the equation.

$2 \sin (x) = \sqrt{2}$

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

Divide each term in $2 \sin (x) = \sqrt{2}$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

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

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

Simplify the right side.

The exact value of $\arcsin (\frac{\sqrt{2}}{2})$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

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.

$x = π - \frac{π}{4}$

Simplify $π - \frac{π}{4}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 - π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π - π}{4}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{4}$

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

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

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

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

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

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

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

Solve for x 2sin(x)cos(x) = square root of 2cos(x)

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