What is the general solution for sin^2(θ)-1=0,0<= θ<= 2pi ?

The general solution for sin^2(θ)-1=0,0<= 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 >

sin^2(θ)-1=0,0<= θ<= 2pi

sin^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Solution steps

sin^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

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 (3 θ) = \frac{3}{2}

tan^{2} (θ) - 1 = 0, π \leq θ \leq 2 π

sin (y) = - \frac{1}{2}

cos^{22} (x) - 2 sin^{2} (x) - 1 = 0

tan (θ) = \frac{3 5}{2}

What is the general solution for sin^2(θ)-1=0,0<= θ<= 2pi ?
The general solution for sin^2(θ)-1=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2