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Trigonometric equation

Posted November 2nd, 2009 by alpha
in

Solve for x:
$ sin^2x(1+tanx)=3sinx(cosx-sinx)+3 $

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Comments

answer

On November 2nd, 2009 Structure says:

$ sin^2x(1+tanx)=3sinx(cosx-sinx)+3=3sinx(cosx-sinx)+3(\sin^2x+\cos^2x) $
Divide by $ \cos^2x $
you get an equation in $ t=\tan x $

$$t^2(1+t)=3t(1-t)+3(t^2+1)$$

or

$$t^3+t^2-3t-3=0$$
$$(t^2-3)(t+1)=0$$
$$t=\tan x=1\:,x=k\pi+\frac{\pi}{4}$$
$$t=\tan x=\sqrt3\:,x=k\pi+\frac{\pi}{3}$$
$$t=\tan x=-\sqrt3\:,x=k\pi-\frac{\pi}{3}$$

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