How can i find x?

(2-cosx/1.341)^2(-sinx/1.341)^2=1

1 Answer
Manikandan S.
Dec 29, 2017

(2-(cosx)/(1.341))^2((-sinx)/1.341)^2 = 1
(2\times 1.341 - cosx)^2(sin^2x) = 1\times(1.341)^4
(7.1931 - 5.364 cos x + cos^2x)(sin^2x) = 3.2338
Let us use sin^2x + cos^2x = 1 as sin^2x = 1-cos^2x. Then
(7.1931 - 5.364 cos x + cos^2x)(1-cos^2x) = 3.2338
(-cos^4x + 5.364 cos^3x -7.1931cos^2x + cos^2x -5.364 cosx +7.1931 - 3.2338) = 0
(cos^4x - 5.364 cos^3x +6.1931cos^2x+5.364 cosx -3.9593) = 0

Let y = cosx then we have a fourth order polynomial. Solving this leads to 4 solutions for cos x You can find the x as cos^(-1) y. I have the solutions as

y = 2.8418 \pm 0.6332 i, -0.8617, 0.5421

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