Find All Complex Solutions 2sin(x)+cos(2x)=1

$2 \sin (x) + \cos (2 x) = 1$

Step 1

Subtract $1$ from both sides of the equation.

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

Step 2

Simplify the left side of the equation.

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Combine the opposite terms in $2 \sin (x) + 1 - 2 \sin^{2} (x) - 1$ .

Subtract $1$ from $1$ .

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

Add $2 \sin (x)$ and $0$ .

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

Step 3

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

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

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

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

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

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

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

Step 4

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

$\sin (x) = 0$

$1 - \sin (x) = 0$

Step 5

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

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

$\sin (x) = 0$

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

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

$x = \arcsin (0)$

Simplify the right side.

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

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.

$x = π - 0$

Subtract $0$ from $π$ .

$x = π$

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.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 6

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

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

$1 - \sin (x) = 0$

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

Subtract $1$ from both sides of the equation.

$- \sin (x) = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$\sin (x) = 1$

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

$x = \arcsin (1)$

Simplify the right side.

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

$x = \frac{π}{2}$

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.

$x = π - \frac{π}{2}$

Simplify $π - \frac{π}{2}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

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.

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

Step 7

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

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

Step 8

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

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

Find All Complex Solutions 2sin(x)+cos(2x)=1

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