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

1. Azteck

$3\sin^{2}(2x)= 3(1-\cos^{2}(2x))$ $=3-3\cos^2(2x)$ ---------------------------------------------------------------------- $3-3\cos^2(2x)+2cos(2x)-2=0$ 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.

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

oh wait just got it!

4. Azteck

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

5. Azteck

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).

6. Azteck

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.

7. Azteck

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

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

wait i'll try and calculate the angle

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

11. Azteck

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.

12. Azteck

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.