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solve $\cos 2x - 5\sin x + 2 = 0, 0^{\circ} \le x \le 360^{\circ}$

I am going to use this formula for $\cos 2x$

$\cos 2x = 1 - 2\sin^2 x$

=> $1 - 2\sin^2x - 5\sin x + 2 = 0$

=> $2\sin^2x - 5\sin x + 3 = 0$

$\sin x(2x - 3)(x -1) =0$

From here I get:

$x = \arcsin -\frac{3}{2}$

and

$x = \arcsin -1$

Could I please have answers using addition formulae and not the unit circle as I have not covered that yet.

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    $\begingroup$ $\cos 2x=1-\color{red}{2}\sin^2 x$. $\endgroup$ – mathlove Apr 12 '16 at 5:50
  • $\begingroup$ @mathlove thank you, I have updated the question $\endgroup$ – dagda1 Apr 12 '16 at 6:00
  • $\begingroup$ $(2\sin x -3)(\sin x -1)=0$ $\endgroup$ – Roman83 Apr 12 '16 at 6:01
  • $\begingroup$ But $2\sin x -3 \not = 0$. Then $\sin x=1$ $\endgroup$ – Roman83 Apr 12 '16 at 6:02
  • $\begingroup$ $2\sin^2x-5\sin x+3=0$ is wrong. $2\sin ^2x\color{red}{+}5\sin x\color{red}{-}3=0$. $\endgroup$ – mathlove Apr 12 '16 at 6:05
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Your factorization is wrong. From the line $2sin^2x-5sinx+3=0$ you have $$u= sinx$$ $$2sin^2x-5sinx+3=0 \rightarrow 2u^2-5x+3=0$$ $$(2u-3)(u-1)=0$$ $$2u-3=0 \rightarrow u= \frac{3}{2}$$ $$sinx= \frac{3}{2} \rightarrow x \in C$$ $$u-1=0 \rightarrow sinx=1$$ $$x= 90^o$$

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  • $\begingroup$ what does $sinx= \frac{3}{2} \rightarrow x \in C$ mean? $\endgroup$ – dagda1 Apr 12 '16 at 6:03
  • $\begingroup$ also why does $u = \sin x$ $\endgroup$ – dagda1 Apr 12 '16 at 6:04
  • $\begingroup$ C is the complex numbers. An imaginary argument could get sin(x) >1. u= sin(x) is just a substitution to make factorization more obvious. I was not sure in the OP if sinx(2x-3)(x-1) was meant to be sin(x)*(2x-3)*(x-1) or if the factorization was (2x-3)(x-1) where x is really sin(x). Using x to replace sin(x) was confusing. $\endgroup$ – Mark Apr 12 '16 at 6:05
  • $\begingroup$ $2\sin ^2x-5\sin x+3=0$ is wrong. $\endgroup$ – mathlove Apr 12 '16 at 6:07
  • $\begingroup$ are you saying that $sinx = \frac{3}2$ is not a valid answer and x = 90 is the only valid answer. Would it be possible to update your answer? I am very interested to know. $\endgroup$ – dagda1 Apr 12 '16 at 6:07

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