I want to solve the equation $\sin(3x)\cos(x)=\frac{2}{3}$ for $x\in [0,\frac{\pi}{2})$ but I could not do it. I tried to develop $\sin(3x)=\sin(x+x+x)$ and I arrived at the step where the equation becomes $(4\cos^2(x)-1)\sin(x)\cos(x)=\frac{2}{3}$ and I could not go further !



$$2=3\sin x\cos x(4\cos^2x-1)$$

Divide both sides by $\cos^4x$

$$3\tan x(3-4\tan^2x)=2\sec^4x=2(1+\tan^2x)^2$$

$$2t^4+12t^3+4t^2-9t+2=0$$ where $t=\tan x$

  • $\begingroup$ $2=3\sin x\cos x(4\cos^2x-3)$ surely? $\endgroup$ – Lord Shark the Unknown Apr 27 '18 at 9:54
  • $\begingroup$ @lab bhattacharjee I think it is $3t^3$ instead of $12t^3$. $\endgroup$ – palio Apr 27 '18 at 9:56
  • $\begingroup$ @LordSharktheUnknown, $$2=3\cos x(3\sin x-4\sin^3x)=3\sin x\cos x(3-4\sin^2x)$$ right? $\endgroup$ – lab bhattacharjee Apr 27 '18 at 9:59
  • $\begingroup$ @palio, Why? Are you sure? $\endgroup$ – lab bhattacharjee Apr 27 '18 at 10:00

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