\text{Solve for }x of \sin^{2}{\left ( x \right )} = \cos{\left ( x \right )}

x=\left\{ \begin{array}{l}-\arccos\left(\frac{-1+\sqrt{5}}{2}\right)+2k\pi \\\arccos\left(\frac{-1+\sqrt{5}}{2}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}

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

Evaluate

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

Use sin^{2} (x) = 1 - cos^{2} (x) to rewrite the expression

1 - cos^{2} (x) = cos (x)

Move the expression to the left side

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

Rewrite in standard form

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

Multiply both sides

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

Substitute a= 1,b= 1 and c= - 1 into the quadratic formula cos (x) = \frac{- b \pm b ^{2} - 4 a c}{2 a}

cos (x) = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 )}{2}

Simplify the expression

More Steps

cos (x) = \frac{- 1 \pm 5}{2}

Separate the equation into 2 possible cases

cos (x) = \frac{- 1 + 5}{2} cos (x) = \frac{- 1 - 5}{2}

Use \frac{- a}{b} = \frac{a}{- b} = - \frac{a}{b} to rewrite the fraction

cos (x) = \frac{- 1 + 5}{2} cos (x) = - \frac{1 + 5}{2}

Rearrange the terms

cos (x) = \frac{- 1 + 5}{2} x \in / R

Calculate

More Steps

x = ⎩ ⎨ ⎧ - arccos (\frac{- 1 + 5}{2}) + 2 kπ arccos (\frac{- 1 + 5}{2}) + 2 kπ, k \in Z x \in / R

Solution

x = ⎩ ⎨ ⎧ - arccos (\frac{- 1 + 5}{2}) + 2 kπ arccos (\frac{- 1 + 5}{2}) + 2 kπ, k \in Z

Alternative Form

x \approx {- 51.82729 2^{\circ} + 36 0^{\circ} k 51.82729 2^{\circ} + 36 0^{\circ} k, k \in Z

Alternative Form

x \approx {- 0.904557 + 2 kπ 0.904557 + 2 kπ, k \in Z

sin^2 ( x ) = cos ( x )

Solve the equation

Graph

Frequently Asked Questions

$Solve for x$ of \(\sin^{2}{\left ( x \right )} = \cos{\left ( x \right )}\)

sin^2 ( x ) = cos ( x )

Solve the equation

Graph

Frequently Asked Questions

Solve for x of \(\sin^{2}{\left ( x \right )} = \cos{\left ( x \right )}\)

$Solve for x$ of \(\sin^{2}{\left ( x \right )} = \cos{\left ( x \right )}\)