Question

Can someone PLEASE help me with Trigonometry???

I can't seem to figure this one out!

Find all solutions if 0° ≤ θ < 360°. When necessary, round your answers to the nearest tenth of a degree. (Enter your answers as a comma-separated list.)

sin^(2) 2θ − 7 sin 2θ − 1 = 0

Answer 1

`\bf sin^2(2\theta)-7sin(2\theta)-1=0\\\\ -------------------------------\\\\ sin(2\theta)=\cfrac{7\pm√(49-4(1)(-1))}{2(1)}\implies sin(2\theta)=\cfrac{7\pm√(49+4)}{2} \\\\\\ sin(2\theta)=\cfrac{7\pm√(53)}{2}\implies 2\theta=sin^(-1)\left( (7\pm√(53))/(2) \right) \\\\\\ \theta=\cfrac{sin^(-1)\left( (7\pm√(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( (7\pm√(53))/(2) \right)}{2}\\\\ -------------------------------\\\\ sin^(-1)\left( (7+√(53))/(2) \right)\implies sin^(-1)(7.14)\impliedby \textit{greater than 1, no good} \\\\\\ sin^(-1)\left( (7-√(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