Trigonometry Can Be A Pain
Algebra
Level
2
If we are given the equation (sin x/sin 2x)/sin 3x =1 Which MUST be true for all solutions?
sin x=2 sin x cos x(cos x - sin^2 x+1/2)
sin x=2 sin x cos x(2 cos^x - sin^2 x)
sin x =2 sin^2 x cos^2 x(3 cos^2 x - sin^2 x)
sin x=2 sin^2 x cos x(3 cos^2 x - sin^2 x)
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1 solution
We can rearrange to get sin x=sin 2x sin 3x. We can further manipulate to get sin x=2 sin x cos x(sin 2x+x).We can change the sin 2x +x term to sin 2x cos x+cos 2x sin x.Going deeper,we get 2 sin cos^2 x + cos^2 x sin x -sin^3 x.We can factor out sin x to get sin x(2 cos^2 x +cos^2 x-sin^2 x) which is basically sin x(3 cos^2 x -sin^2 x).Adding in the other terms,we get sin x =2 sin^2 cos x(3 cos^2 x -sin^2 x)