trigonometry
Find the maximum angle $X$ in the range $0^\circ \leq x < 360^\circ$ which satisfies the equation
$\cos^2(2x)+\sqrt{3}\sin(2x)-\frac{7}{4}=0$
$\cos^2(2x)=\sin^2(2x)-1$, so we can substitute $t$ for $\sin(2x)$, and we have the equation $t^2+\sqrt{3}t-\frac{7}{4}-1=0$.
Solving for $t$, $t$ is either $\frac{-\sqrt{3}+\sqrt{14}}{2}$ or $\frac{-\sqrt{3}-\sqrt{14}}{2}$
$x= \frac{\arcsin(t)}{2}$
How can I find the maximum angle $X$?
Edit: $\cos^2(2x) = -\sin^2(2x)+1$ , not $\cos^2(2x)=\sin^2(2x)-1$
A slightly 'expanded-upon' version of user67418's answer:
The circle here represents the parametric curve $(x=\cos\theta, y=\sin\theta)$, and the line is the line $x+y=1$, so their points of intersection are the points where $\cos\theta+\sin\theta=1$; at least for me, this is the clearest way of seeing that there are only the two solutions already mentioned.
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$?
It is the latter.
So, having $\tan\alpha=-6/8=-3/4$ gives $\alpha=\arctan(-3/4)\approx -36.9^\circ$.
Hence, for $n\in\mathbb Z$, $$\begin{align}10\cos(x-(-36.9^\circ))=5&\Rightarrow \cos(x+36.9^\circ)=1/2\\&\Rightarrow x+36.9^\circ=\pm 60^\circ+360^\circ n\\&\Rightarrow x=-36.9^\circ\pm 60^\circ+360^\circ n\\&\Rightarrow x=23.1^\circ,\quad 263.1^\circ\end{align}$$ since $0^\circ\le x\le 360^\circ$.
Best Answer
Note you have $\cos^22x=1-\sin^22x$; with $\sin2x=t$ you get $$ 1-t^2+t\sqrt{3}-\frac{7}{4}=0 $$ so the equation can be rewritten $$ 4t^2-4t\sqrt{3}+3=0 $$ This is actually $(2t-\sqrt{3})^2=0$, so the only root is $t=\sqrt{3}/2$. Hence $2x=\pi/3+2k\pi$ or $2x=2\pi/3+2k\pi$. Hence $$ x=\frac{\pi}{6}+k\pi\qquad\text{or}\qquad x=\frac{\pi}{3}+k\pi $$ If you prefer degrees, $$ x=30^\circ+k180^\circ\qquad\text{or}\qquad x=60^\circ+k180^\circ $$ It shouldn't be difficult to find the maximum solution.