What is the general solution for sin^4(x)+cos^4(x)=cos^4(x) ?

The general solution for sin^4(x)+cos^4(x)=cos^4(x) is x=2pin,x=pi+2pin

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sin^4(x)+cos^4(x)=cos^4(x)

sin^{4} (x) + cos^{4} (x) = cos^{4} (x)

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{4} (x) + cos^{4} (x) = cos^{4} (x)

Subtract

cos^{4} (x)

from both sides

sin^{4} (x) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

sin (x) = 0

The solution is

sin (x) = 0

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

cot (x) = \frac{40}{32}

tan (a) = \frac{3}{8}

sin (2 x + 10) = cos (3 x - 20)

sin^{3} (x) = - \frac{2}{3}

x, tan^{2} (x) = tan^{2} (y)

What is the general solution for sin^4(x)+cos^4(x)=cos^4(x) ?
The general solution for sin^4(x)+cos^4(x)=cos^4(x) is x=2pin,x=pi+2pin