Solve the equation :
$$\sin 3x=2\cos^3x$$
my try :
$\sin 3x=3\sin x-4\sin^3x$
$\cos^2x=1-\sin^2x$
so:
$$3\sin x-4\sin^3x=2((1-\sin^2x)(\cos x))$$
then ?
Solve the equation: $\sin 3x=2\cos^3x$
1
$\begingroup$
trigonometry
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0You have too many `2`s on the right hand side. – 2017-01-24
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0And I don't think the left hand side is very helpful. I would look for ways to reduce everything in terms of $sin x$ and $cos x$ – 2017-01-24
2 Answers
5
$$2\cos^3x=3\sin x-4\sin^3x$$
Divide both sides by $\cos^3x$
$$2=3\tan x(1+\tan^2x)-4\tan^3x$$
$$\tan^3x-3\tan x+2=0$$ which is a cubic equation in $\tan x$
Clearly, one of the roots is $\tan x=1$
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1Oh, even better, a quadratic in tan! :D – 2017-01-24
6
Hint:
$$\sin 3x=2\cos^3x$$ $$3\sin x-4\sin^3x=2\cos^3x$$ $$3\sin x(\cos^2x+\sin^2x)-4\sin^3x=2\cos^3x$$ $$\sin^3x-3\sin x\cos^2x+2\cos^3x=0$$ Divide both sides by $\cos^3x$ and $\tan x=t$ $$t^3-3t+2=0$$ $$(t-1)^2(t+2)=0$$