4cos2x = 8sinxcosx

V

Vol

Junior Member
Joined
May 23, 2018
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Hi, y'all. I can't figure out what trig identities were used.
4cos2x = 8sinxcosx
4cos2x - 8sinxcosx = 0
Now I'm stuck. I know the answer is 4 - 4tan2x=0 but how what it done?
Thanks.
 
What exactly are you trying to do? The statement itself is NOT an identity, it's not true for x= 0. If you are asking for an identity that will help solve the equation? Then, yes, sin2x= 2 sin(x)cos(x) so 8sin(x)cos(x)= 4(2sin(x)cos(x))= 4sin 2x. So the equatioin can be written as 4 cos(2x)= 4 sin(2x). cos(2x)= sin(2x). Dividing both sides by cos(2x), sin(2x)/cos(2x)= tan(2x)= 1. tangent is equal to 1 when \(\displaystyle x= \pi/4\), \(\displaystyle x= -\pi/4\), and multiples of \(\displaystyle 2\pi\) added to that. So \(\displaystyle 2x= \frac{\pi}{4}+ 2n\pi\) and \(\displaystyle 2x= -\frac{\pi}{4}+ 2n\pi\) so that \(\displaystyle x= \frac{\pi}{8}+ n\pi\) and \(\displaystyle x= -\frac{\pi}{8}+ n\pi\).​
 
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Oh, oops. The problem says solve for x belonging to 0 and 2pi.
 
I couldn't see you had to divide both sides by cos2x to turn it into tan2x, which looks nicer. Thanks y'all.
 
OK, so you did not see dividing by cos(2x).
You should know when sin(x) = cos(x) for 0<x<2pi
So how would you solve sin(2x) = cos(2x) for 0<x<2pi?
 
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