What is the general solution for -sin^2(x)=cos(2x) ?

The general solution for -sin^2(x)=cos(2x) is x= pi/2+2pin,x=(3pi)/2+2pin

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

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

cos (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Simplify

= - cos^{2} (x)

- cos^{2} (x) = 0

Divide both sides by

- 1

- cos^{2} (x) = 0

Divide both sides by

- 1

- cos^{2} (x) = 0

Divide both sides by

- 1

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

Simplify

cos^{2} (x) = 0

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{4}{5}, \frac{π}{2} \leq 0 < π, sin (2 x)

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, \frac{3 π}{2} < θ < 2 π

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

3 cos (x) - 2 sin (2 x) = 0

4 cos (2 t) = 2

What is the general solution for -sin^2(x)=cos(2x) ?
The general solution for -sin^2(x)=cos(2x) is x= pi/2+2pin,x=(3pi)/2+2pin