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

The general solution for cos^2(θ)+2cos(θ)+1=0,0<= θ<= 2pi is θ=pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(θ)+2cos(θ)+1=0,0<= θ<= 2pi

cos^{2} (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

θ = π

Solution steps

cos^{2} (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

Solve by substitution

cos (θ) = - 1

cos (θ) = - 1, 0 \leq θ \leq 2 π : θ = π

Combine all the solutions

θ = π

cot (θ) - 3 = 0, 0 \leq θ \leq 2 π

2 = sec (θ)

sin (x) = \frac{1}{8}, sin (2 x)

x, csc (x) = 2

\frac{3}{2} sec (x) = 3

What is the general solution for cos^2(θ)+2cos(θ)+1=0,0<= θ<= 2pi ?
The general solution for cos^2(θ)+2cos(θ)+1=0,0<= θ<= 2pi is θ=pi