Solve for x sin(2x)=2cos(x)^2

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

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

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

Apply the sine double-angle identity.

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

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

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

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

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

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

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

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

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

$\cos (x) = 0$

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

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

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

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

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

Divide each term in the equation by $\cos (x)$ .

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

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$\tan (x) + \frac{- \cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Cancel the common factor of $\cos (x)$ .

Cancel the common factor.

$\tan (x) + \frac{- \cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Divide $- 1$ by $1$ .

$\tan (x) - 1 = \frac{0}{\cos (x)}$

Separate fractions.

$\tan (x) - 1 = \frac{0}{1} \cdot \frac{1}{\cos (x)}$

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$\tan (x) - 1 = \frac{0}{1} \cdot \sec (x)$

Divide $0$ by $1$ .

$\tan (x) - 1 = 0 \sec (x)$

Multiply $0$ by $\sec (x)$ .

$\tan (x) - 1 = 0$

Add $1$ to both sides of the equation.

$\tan (x) = 1$

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

$x = \arctan (1)$

Simplify the right side.

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth 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 $π$ .

$x = \frac{4 \cdot π + π}{4}$

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

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

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

Consolidate the answers.

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

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

Consolidate $\frac{π}{4} + π n$ and $\frac{5 π}{4} + π n$ to $\frac{π}{4} + π n$ .

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

Solve for x sin(2x)=2cos(x)^2

Related Questions

Support LCMGCF.com by disabling your adblocker.