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Number of solution of the equation

$\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4\; \forall $ $x\in(0,2\pi)$

what i try

$\cos^4(2x)+2\sin^2 2x=17(1+\sin^2(2x)+2\sin 2x)^2$

$1+\sin^4 (2x)=17(1+\sin^4 2x+2\sin^2 2x+4\sin^24x+4\sin 2x(1+\sin^2 2x))$

$16\sin^4 (2x)+68\sin^3 2x+34\sin^2 2x+68\sin 2x+68\sin^2 4x+16=0$

How do i solve it Help me please

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Hint: Your equation is equivalent to $$2 (\sin (2 x)+2) (2 \sin (2 x)+1) (-7 \sin (2 x)+2 \cos (4 x)-6)=0$$

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  • $\begingroup$ And how did you find this factorization? $\endgroup$ – lhf Jul 22 at 13:49
  • $\begingroup$ A little bit of intuition, and i have solved many of these similar problems in our math camp in Leipzig. $\endgroup$ – Dr. Sonnhard Graubner Jul 22 at 13:55
  • $\begingroup$ @Dr.SonnhardGraubner That doesn't tell me anything though. $\endgroup$ – Sudix Jul 22 at 17:53
  • $\begingroup$ You must set the factors equal to zero! $\endgroup$ – Dr. Sonnhard Graubner Jul 22 at 17:57
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You're not required to find all solutions, just to find how many there are.

Let $u=\sin(2x)$. Then the trigonometric equation in $x$ becomes a polynomial equation in $u$: $$ 0 = (1 - u)^4 + 2 u^2 - 17 (1 + u)^4 = -2 (8 u^4 + 36 u^3 + 47 u^2 + 36 u + 8) $$ Now plot this function of $u$ and see how many solutions are in the interval $[-1,1]$ so that $u$ can be the sine of something.

The graph below tells us that there is exactly one solution in $[-1,1]$.

enter image description here

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Wrong expansion jacky $$16\sin^42x+68\sin^32x+102\sin^22x+68\sin 2x+16=0$$ $$16(\sin 2x+\csc 2x)^2+68(\sin 2x+\csc 2x)+70=0$$ $$\sin 2x +\csc 2x=-15.0479,-1.9521$$ Rejecting $-1.9521$ (Why)

Thus, $$\sin^22x+(15.0479)\sin 2x+1=0$$ Both the roots of the quadratic are negative and since their product is $1$ only one root will lie on the interval $(-1,0)$

So there should be $4$ solutions

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