Question

For the following exercises, solve with the methods shown in this section exactly on the interval $[0, 2\pi)$. 33. $\sin(3x)\cos(6x) - \cos(3x)\sin(6x) = -0.9$ 35. $\cos(2x)\cos x + \sin(2x)\sin x = 1$ 37. $9\cos(2\theta) = 9\cos^2 \theta - 4$ 39. $\cos(2t) = \sin t$ 34. $\sin(6x)\cos(11x) - \cos(6x)\sin(11x) = -0.1$ 36. $6\sin(2t) + 9\sin t = 0$ 38. $\sin(2t) = \cos t$ 40. $\cos(6x) - \cos(3x) = 0$

          For the following exercises, solve with the methods shown in this section exactly on the interval $[0, 2\pi)$. 
33. $\sin(3x)\cos(6x) - \cos(3x)\sin(6x) = -0.9$
35. $\cos(2x)\cos x + \sin(2x)\sin x = 1$
37. $9\cos(2\theta) = 9\cos^2 \theta - 4$
39. $\cos(2t) = \sin t$
34. $\sin(6x)\cos(11x) - \cos(6x)\sin(11x) = -0.1$
36. $6\sin(2t) + 9\sin t = 0$
38. $\sin(2t) = \cos t$
40. $\cos(6x) - \cos(3x) = 0$

For the following exercises, solve with the methods shown in this section exactly on the interval [0, 2π).
33. sin(3x)cos(6x) - cos(3x)sin(6x) = -0.9
35. cos(2x)cos x + sin(2x)sin x = 1
37. 9cos(2θ) = 9cos^2 θ - 4
39. cos(2t) = sin t
34. sin(6x)cos(11x) - cos(6x)sin(11x) = -0.1
36. 6sin(2t) + 9sin t = 0
38. sin(2t) = cos t
40. cos(6x) - cos(3x) = 0

Added by Jodi B.

Precalculus with Limits

Ron Larson 2nd Edition

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For the following exercises, solve with the methods shown in this section exactly on the interval [0,2pi ). 40. cos(6x)-cos(3x)=0 For the following exercises,solve with the methods shown in this section exactly on the interval [0,2n) 33.sin3xcos6x-cos3xsin6x=-0.9 34.sin6xcos11x-cos6xsin11x=-0.1 35.cos2xcosx+sin2xsinx=1 36.6sin2t+9sint=0 37.9cos20=9cos-4 38.sin2t=cost 39.cos2t=sin t 40.cos6xcos3x=0