Need help with a math problem. Any help is greatly appreciated.
Find all degree solutions in the interval $0° ≤ θ < 360°$. If rounding is necessary, round to the nearest tenth of a degree. Use your graphing calculator to verify the solution graphically. (Enter your answers as a comma-separated list.)
$5\sin^2 θ − 6 \cos 2θ = 0$
I used the following double angle formula to make the entire equation sin:
$1-2\sin^2θ=0$
$5\sin^2 θ-6(1-2\sin^2θ)=0$
$5\sin^2θ+12\sin^2θ-6=0$
$17\sin^2θ-6=0$
$17\sin^2θ=6$
$\sin^2θ=6/17$
$\sinθ=\pm \sqrt{6/17}$
After taking the inverse of $\sin$ I got
$θ = 36.5°, 130.3°$
Best Answer
you could also do $$0 = 5 \sin^2 t - 6 \cos 2t =\frac 52\left(1-\cos 2t\right) - 6\cos 2t \to \cos 2t = \frac 5{17} $$
solutions are $$2t = \pm \cos^{-1}\left(\frac5{17}\right) + 360^\circ k =\pm 72.8953^\circ+360^\circ k$$ that is $$t = 36.4476, 143.552^\circ, 216.447^\circ,323.552^\circ $$