# solve for x: sinxcosx+0.5=sin^2 x

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• Jun 22nd 2009, 09:06 PM
yoman360
solve for x: sinxcosx+0.5=sin^2 x
did I do this correctly so far?
sinxcosx+0.5=sin^2 x
-sin^2 x + sinxcosx = -0.5
-sin x (sin x - cos x) = -0.5
sin x (sin x - cos x) = 0.5

If I did this correctly so far where do I go from here?
• Jun 22nd 2009, 09:31 PM
Grandad
Hello yoman360
Quote:

Originally Posted by yoman360
did I do this correctly so far?
sinxcosx+0.5=sin^2 x
-sin^2 x + sinxcosx = -0.5
-sin x (sin x - cos x) = -0.5
sin x (sin x - cos x) = 0.5

If I did this correctly so far where do I go from here?

This would be OK if you had zero on the RHS. The best way to go is to use the double-angle formulae: $\sin x\cos x = \tfrac12\sin2x$ and $\sin^2x = \tfrac12(1 - \cos2x)$

$\sin x\cos x+0.5=\sin^2 x$

$\Rightarrow \tfrac12\sin2x +0.5=\tfrac12(1 - \cos2x)$

$\Rightarrow \sin2x +1=1 - \cos2x$

$\Rightarrow \sin 2x +\cos2x =0$

Now use the fact that $\sin(2x + \pi/4) = \sin2x\cos\pi/4 +\cos2x\sin\pi/4 = \tfrac{1}{\sqrt2}(\sin2x+\cos2x)$ and get:

$\sin(2x+\pi/4)=0$

$\Rightarrow 2x +\pi/4 = 0, \pi, 2\pi, ...$

$\Rightarrow x = -\pi/8, 3\pi/8, 7\pi/8, ...$

Grandad
• Jun 23rd 2009, 03:45 AM
Soroban
Hello, yoman360!

Grandad is right . . . We need another approach.

Quote:

$\sin x\cos x+\tfrac{1}{2}\:=\:\sin^2\!x$

We have: . $\sin^2\!x - \sin x\cos x \:=\:\tfrac{1}{2}$

$\text{Multiply by 2: }\;\underbrace{2\sin^2\!x}_{1-\cos2x} - \underbrace{2\sin x\cos x}_{\sin2x} \:=\:1 \quad\Rightarrow\quad 1 - \cos2x - \sin2x \:=\:1$

$\text{We have: }\;\sin2x \:=\:-\cos2x \quad\Rightarrow\quad \frac{\sin 2x}{\cos2x} \:=\:-1$

Hence: . $\tan2x \:=\:-1 \quad\Rightarrow\quad 2x \:=\:\frac{3\pi}{4}+\pi n \quad\Rightarrow\quad \boxed{x \:=\:\frac{3\pi}{8} + \frac{\pi}{2}n}$