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OpenStudy (anonymous):
in terms of what?
OpenStudy (anonymous):
in terms of simplifying? eg. sin2x=2sinxcosx. So what is sin(x/2)?
OpenStudy (anonymous):
well then it will be\[\sin \frac{x}{2}=2 \sin \frac{x}{4} \cos \frac{x}{4}\]am i right?
OpenStudy (anonymous):
ye i hope so :P
OpenStudy (anonymous):
:)
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OpenStudy (anonymous):
so then how would you solve cosx+3sin(x/2) -2 = 0 for 0<=x<=pi?
OpenStudy (anonymous):
i think that will not be useful for this problem, better to write \(\cos x\) in terms of half angles
OpenStudy (anonymous):
so can you explain?
OpenStudy (anonymous):
see, we'r going to make a quadratic out of ur equation
OpenStudy (anonymous):
u sure know the formula\[\cos 2x=\cos^2 x-\sin^2 x\]so\[\cos x=\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}\]
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OpenStudy (anonymous):
so \[\cos x=\color\red{\cos^2 \frac{x}{2}}-\sin^2 \frac{x}{2}=\color\red{1-\sin^2 \frac{x}{2}}-\sin^2 \frac{x}{2}=1-2 \sin^2 \frac{x}{2}\]put this in the equation\[1-2 \sin^2 \frac{x}{2}+3 \sin \frac{x}{2}-2=0\]
OpenStudy (anonymous):
sin pi/2 is 1
OpenStudy (anonymous):
if u let \(u=\sin \frac{x}{2}\) ur equation becomes\[-2u^2+3u-1=0\]which is easy to solve :) makes sense ?