Find All Complex Solutions sin(x)^2=2-2cos(x)

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

Step 1

Move all the expressions to the left side of the equation.

Subtract $2$ from both sides of the equation.

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

Add $2 \cos (x)$ to both sides of the equation.

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

Step 2

Replace $\sin^{2} (x)$ with $1 - \cos^{2} (x)$ .

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

Step 3

Solve for $x$ .

Subtract $2$ from $1$ .

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

Factor by grouping.

Reorder terms.

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

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 1 = 1$ and whose sum is $b = 2$ .

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

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

Rewrite $2$ as $1$ plus $1$

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

Apply the distributive property.

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

Multiply $\cos (x)$ by $1$ .

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

Multiply $\cos (x)$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- \cos (x) + 1$ .

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

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

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

$\cos (x) - 1 = 0$

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

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

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

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

Subtract $1$ from both sides of the equation.

$- \cos (x) = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$\cos (x) = 1$

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

$x = \arccos (1)$

Simplify the right side.

The exact value of $\arccos (1)$ is $0$ .

$x = 0$

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.

$x = 2 π - 0$

Subtract $0$ from $2 π$ .

$x = 2 π$

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.

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

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

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

Step 4

Consolidate the answers.

$x = 2 π n$ , for any integer $n$

Find All Complex Solutions sin(x)^2=2-2cos(x)

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