Question

Can someone PLEASE help me with Trigonometry???

Answer 1

$\bf sin^2(2\theta)-7sin(2\theta)-1=0\\\\ -------------------------------\\\\ sin(2\theta)=\cfrac{7\pm\sqrt{49-4(1)(-1)}}{2(1)}\implies sin(2\theta)=\cfrac{7\pm\sqrt{49+4}}{2} \\\\\\ sin(2\theta)=\cfrac{7\pm\sqrt{53}}{2}\implies 2\theta=sin^{-1}\left( \frac{7\pm\sqrt{53}}{2} \right) \\\\\\ \theta=\cfrac{sin^{-1}\left( \frac{7\pm\sqrt{53}}{2} \right)}{2}$

but anyway, the numerator will give the angles, and θ is just half of each

$\bf \theta=\cfrac{sin^{-1}\left( \frac{7\pm\sqrt{53}}{2} \right)}{2}\\\\ -------------------------------\\\\ sin^{-1}\left( \frac{7+\sqrt{53}}{2} \right)\implies sin^{-1}(7.14)\impliedby \textit{greater than 1, no good} \\\\\\ sin^{-1}\left( \frac{7-\sqrt{53}}{2} \right)\implies sin^{-1}(-0.14) \approx -8.05^o \\\\\\ \theta=\cfrac{-8.05}{2}\implies \theta=-4.025$

ok... that's a negative tiny angle, is in the 4th quadrant, if we stick to the range given, from 0 to 360, so we have to use the positive version of it, 360-4.025

so the angle is 355.975°

now, the 3rd quadrant has another angle whose sine is negative, so... if we move from the 180° line down by 4.025, we end up at 184.025°

and those are the only two angles, because, on the 2nd and 1st quadrants, the sine is positive, so it wouldn't have an angle there