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AzteckBest ResponseYou've already chosen the best response.0
\[3\sin^{2}(2x)= 3(1\cos^{2}(2x))\] \[=33\cos^2(2x)\]  \[33\cos^2(2x)+2cos(2x)2=0\] Collect like terms and move everything to the RHS \[3\cos^2(2x)2cos(2x)1=0\] Let u=cos(2x) \[3u^22u1=0\] Solve the quadratic.
 one year ago

shaqadryBest ResponseYou've already chosen the best response.0
i dont get it, what happen to sin^2 2x?
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Yeah I just made sin^2(2x) in terms of cos(2x).
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
So could you solve that quadratic and then after you have the values of u, bring back the cos(2x) by replacing u with cos(2x).
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Then after you've done that, solve normally as if it were cos(x) and then after you get the angles, divide by 2 because you're finding x, not 2x.
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
If you're unclear about what you're supposed to do, let me know.
 one year ago

shaqadryBest ResponseYou've already chosen the best response.0
okay thanks! i really didnt know how to solve that part bcs it has 2x
 one year ago

shaqadryBest ResponseYou've already chosen the best response.0
wait i'll try and calculate the angle
 one year ago

shaqadryBest ResponseYou've already chosen the best response.0
the angles i got are 0, 54.75, 125.25, 180, 234.75, 305. 25 and 360
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Awesome work. That's pretty much all of them. I think you need to work on your rounding up. Because for some of the angles, I got either 0.01 more or less than your angles.
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Ah I see, you rounded up before dividing by 2. That's okay. You're still correct. I just prefer to round up at the end result.
 one year ago
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