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

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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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

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

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

Prove the identity:

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

1 - 2 sin^{2} (θ) + sin^{4} (θ)

Apply Perfect Square Formula:

a^{2} - 2 ab + b^{2} = (a - b)^{2}

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

= (1 - sin^{2} (θ))^{2}

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

= (cos^{2} (θ))^{2}

Simplify

(cos^{2} (θ))^{2} : cos^{4} (θ)

(cos^{2} (θ))^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= cos^{2 \cdot 2} (θ)

Multiply the numbers:

2 \cdot 2 = 4

= cos^{4} (θ)

= cos^{4} (θ)

= cos^{4} (θ) + cos^{4} (θ)

Simplify

= 2 cos^{4} (θ)

2 cos^{4} (θ) = 0

Divide both sides by

2

2 cos^{4} (θ) = 0

Divide both sides by

2

\frac{2 cos ^{4} ( θ )}{2} = \frac{0}{2}

Simplify

cos^{4} (θ) = 0

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

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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