What is the general solution for 2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi ?

The general solution for 2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi

2 sin^{2} (θ) - 2 = cos^{2} (θ), 0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Solution steps

2 sin^{2} (θ) - 2 = cos^{2} (θ), 0 \leq θ \leq 2 π

Subtract

cos^{2} (θ)

from both sides

2 sin^{2} (θ) - 2 - cos^{2} (θ) = 0

Rewrite using trig identities

- 3 + 3 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1, 0 \leq θ \leq 2 π : θ = \frac{π}{2}

sin (θ) = - 1, 0 \leq θ \leq 2 π : θ = \frac{3 π}{2}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{3 π}{2}

cos (θ) = - \frac{7}{5}

sin (π + x) = sin (x)

cos (x) = - \frac{2}{π}

23 cos (z) + 2 sin (z) = 0

\frac{cos ( x ) + cos ( 3 x )}{cos ^{2} ( x ) - sin ^{2} ( x )} = 0

What is the general solution for 2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi ?
The general solution for 2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2