What is the general solution for sin^4(θ)+cos^4(θ)=2sin^2(θ)-1 ?

The general solution for sin^4(θ)+cos^4(θ)=2sin^2(θ)-1 is θ= pi/2+2pin,θ=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^4(θ)+cos^4(θ)=2sin^2(θ)-1

sin^{4} (θ) + cos^{4} (θ) = 2 sin^{2} (θ) - 1

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

Solution steps

sin^{4} (θ) + cos^{4} (θ) = 2 sin^{2} (θ) - 1

Subtract

2 sin^{2} (θ) - 1

from both sides

sin^{4} (θ) + cos^{4} (θ) - 2 sin^{2} (θ) + 1 = 0

Rewrite using trig identities

2 cos^{4} (θ) = 0

Divide both sides by

2

cos^{4} (θ) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

cos (θ) = 0

The solution is

cos (θ) = 0

General solutions for

cos (θ) = 0

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

x, \frac{- y}{2} + sin (x) = 0

cos (a) = 0.33

tan (θ) = \frac{3}{- 2}

4 sin^{2} (r) - sin (r) = 0

tan (θ) = \frac{3}{- 3}

What is the general solution for sin^4(θ)+cos^4(θ)=2sin^2(θ)-1 ?
The general solution for sin^4(θ)+cos^4(θ)=2sin^2(θ)-1 is θ= pi/2+2pin,θ=(3pi)/2+2pin