Question

sin^2020a-cos^2022=x find the range of x

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Answer 1

Answer:

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

y = (sin^2(x) + cos^2(x))^2 - 2sin^2(x).cos^2(x)

we know from identity : sin^2(x) + cos^2(x) = 1

y = 1 - 2sin^2(x).cos^2(x)

we know from identity that : 2sin(x).cos(x) = sin(2x)

y = 1 - (1/2)[sin(2x)]^2

for max value of y second term must be min. which is when sin(2x) is 0, so max value is 1.

for min value of y second term must be max. which is when sin(2x) is 1, so min value is 1/2

therrfore range of the above equation is : [1/2 , 1]