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Trigonometry equations?
I shall be grateful to anyone who can help me with the following equation.
4(cos^6x + sin^6x) = 1 + 2cos^2 2x
1 Answer
- JBLv 71 decade agoFavorite Answer
It is not an identity so presumably you want to find values of x that make it true, i.e. solve the equation. The left side is 4 times a sum of two cubes so it factors as:
4(sin^2 x + cos^2 x)(cos^4 x - sin^2 x cos^2 x + sin^4 x) and the first factor after the 4 equals 1. So your equation is now:
4(cos^4 x - sin^2 x cos^2 x + sin^4 x) - 1 - 2 cos^2 2x = 0
Now sin^2 x = 1-cos^2 x and sin^4 x = (sin^2 x )^2 = (1-cos^2 x)^2 = 1 - 2cos^2 x + cos^4 x and cos^2 2x = (2cos^2 x - 1)^2 = 4cos^4 x - 4cos^2 x + 1 so substituting and simplifying gives:
4 cos^4 x - 4 cos^2 x + 1 = 0 which factors:
(2 cos^2 x - 1)^2 = 0
You take it from there.
