# solve $\cos 2x - 5\sin x + 2 = 0$ using addition formulae

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.

• $\cos 2x=1-\color{red}{2}\sin^2 x$. – mathlove Apr 12 '16 at 5:50
• @mathlove thank you, I have updated the question – dagda1 Apr 12 '16 at 6:00
• $(2\sin x -3)(\sin x -1)=0$ – Roman83 Apr 12 '16 at 6:01
• But $2\sin x -3 \not = 0$. Then $\sin x=1$ – Roman83 Apr 12 '16 at 6:02
• $2\sin^2x-5\sin x+3=0$ is wrong. $2\sin ^2x\color{red}{+}5\sin x\color{red}{-}3=0$. – mathlove Apr 12 '16 at 6:05

Your factorization is wrong. From the line $2sin^2x-5sinx+3=0$ you have $$u= sinx$$ $$2sin^2x-5sinx+3=0 \rightarrow 2u^2-5x+3=0$$ $$(2u-3)(u-1)=0$$ $$2u-3=0 \rightarrow u= \frac{3}{2}$$ $$sinx= \frac{3}{2} \rightarrow x \in C$$ $$u-1=0 \rightarrow sinx=1$$ $$x= 90^o$$
• what does $sinx= \frac{3}{2} \rightarrow x \in C$ mean? – dagda1 Apr 12 '16 at 6:03
• also why does $u = \sin x$ – dagda1 Apr 12 '16 at 6:04
• $2\sin ^2x-5\sin x+3=0$ is wrong. – mathlove Apr 12 '16 at 6:07
• 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. – dagda1 Apr 12 '16 at 6:07