What is the general solution for -2sin^2(θ)+cos(θ)=-1 ?

The general solution for -2sin^2(θ)+cos(θ)=-1 is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-2sin^2(θ)+cos(θ)=-1

- 2 sin^{2} (θ) + cos (θ) = - 1

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = π + 2 πn

Solution steps

- 2 sin^{2} (θ) + cos (θ) = - 1

Subtract

- 1

from both sides

- 2 sin^{2} (θ) + cos (θ) + 1 = 0

Rewrite using trig identities

- 1 + cos (θ) + 2 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = π + 2 πn

4 cos (θ) = 4 (1 - cos (θ))

2 cos (x) = 1.7

sin (2 x + 30) = 0.8

1 - 2 sin (x) = - 1

4 tan^{2} (x) + sec (x) = - 2 tan^{2} (x) + sec^{2} (x) - 3

What is the general solution for -2sin^2(θ)+cos(θ)=-1 ?
The general solution for -2sin^2(θ)+cos(θ)=-1 is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=pi+2pin