# Given that sin^4(x) + cos^4(x) = 1, find all values for x between 0 and 2π

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### 1 Answer

(sinx)^4+(cosx)^4 = 1

The soltion is obvious but a procedure is as below:

s^4+(1-s^2)^2 = 1, where s = sinx. Also (sinx)^2+(cosx)^2 = 1 is a trigonometric identity for all x. So, (cosx)^4 = [(cosx)^2]^2 = (1-s^2)^2. Therefore,

s^4+1-2s^2+s^4=1

2s^4-2s^2=0

s^4-s^2 = 0

s^2(s^2-1)=0

s^2 = 0 or s^2 = 1

s=0 or s=1 or s=-1

sinx = 0 or sinx = 1 or sin x= -1

When sinx = o, x = 0 degree or x=180 derees.

When sinx = 1, x= 90 degree

When sinx = -1, x= 270 degree.