Question

how to solve 3 sin^2 2x + 2 cos 2x - 2 = 0

Answer 1

\[3\sin^{2}(2x)= 3(1-\cos^{2}(2x))\]
\[=3-3\cos^2(2x)\]
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\[3-3\cos^2(2x)+2cos(2x)-2=0\]
Collect like terms and move everything to the RHS
\[3\cos^2(2x)-2cos(2x)-1=0\]

Let u=cos(2x)
\[3u^2-2u-1=0\]
Solve the quadratic.

Answer 2

i dont get it, what happen to sin^2 2x?

Answer 3

oh wait just got it!

Answer 4

Yeah I just made sin^2(2x) in terms of cos(2x).

Answer 5

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).

Answer 6

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.

Answer 7

If you're unclear about what you're supposed to do, let me know.

Answer 8

okay thanks! i really didnt know how to solve that part bcs it has 2x

Answer 9

wait i'll try and calculate the angle

Answer 10

the angles i got are 0, 54.75, 125.25, 180, 234.75, 305. 25 and 360

Answer 11

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.

Answer 12

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.