Solve for x 4cos(x)=-sin(x)^2+1

$4 \cos (x) = - \sin^{2} (x) + 1$

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

Add $\sin^{2} (x)$ to both sides of the equation.

$4 \cos (x) + \sin^{2} (x) = 1$

Subtract $1$ from both sides of the equation.

$4 \cos (x) + \sin^{2} (x) - 1 = 0$

Simplify $4 \cos (x) + \sin^{2} (x) - 1$ .

Reorder $\sin^{2} (x)$ and $- 1$ .

$4 \cos (x) - 1 + \sin^{2} (x) = 0$

Rewrite $- 1$ as $- 1 (1)$ .

$4 \cos (x) - 1 (1) + \sin^{2} (x) = 0$

Factor $- 1$ out of $\sin^{2} (x)$ .

$4 \cos (x) - 1 (1) - 1 (- \sin^{2} (x)) = 0$

Factor $- 1$ out of $- 1 (1) - 1 (- \sin^{2} (x))$ .

$4 \cos (x) - 1 (1 - \sin^{2} (x)) = 0$

Rewrite $- 1 (1 - \sin^{2} (x))$ as $- (1 - \sin^{2} (x))$ .

$4 \cos (x) - (1 - \sin^{2} (x)) = 0$

Apply pythagorean identity.

$4 \cos (x) - \cos^{2} (x) = 0$

Solve for $x$ .

Factor the left side of the equation.

Let $u = \cos (x)$ . Substitute $u$ for all occurrences of $\cos (x)$ .

$4 u - u^{2} = 0$

Factor $u$ out of $4 u - u^{2}$ .

Factor $u$ out of $4 u$ .

$u \cdot 4 - u^{2} = 0$

Factor $u$ out of $- u^{2}$ .

$u \cdot 4 + u (- u) = 0$

Factor $u$ out of $u \cdot 4 + u (- u)$ .

$u (4 - u) = 0$

Replace all occurrences of $u$ with $\cos (x)$ .

$\cos (x) (4 - \cos (x)) = 0$

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

$\cos (x) = 0$

$4 - \cos (x) = 0$

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 $4 - \cos (x)$ equal to $0$ and solve for $x$ .

Set $4 - \cos (x)$ equal to $0$ .

$4 - \cos (x) = 0$

Solve $4 - \cos (x) = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$- \cos (x) = - 4$

Divide each term in $- \cos (x) = - 4$ by $- 1$ and simplify.

Divide each term in $- \cos (x) = - 4$ by $- 1$ .

$\frac{- \cos (x)}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\cos (x)}{1} = \frac{- 4}{- 1}$

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

$\cos (x) = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$\cos (x) = 4$

The range of cosine is $- 1 \leq y \leq 1$ . Since $4$ does not fall in this range, there is no solution.

No solution

The final solution is all the values that make $\cos (x) (4 - \cos (x)) = 0$ true.

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

Consolidate the answers.

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

Solve for x 4cos(x)=-sin(x)^2+1

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