# If $\sin32^{\circ}=k$ and $\cos x=1-2k^2, \alpha$ and $\beta$ are the two values of $x$ between $0^{\circ}$ and $360^{\circ}$ with $\alpha<\beta$

A)$$\alpha+\beta=180^{\circ}$$

B)$$\beta-\alpha=200^{\circ}$$

C)$$\beta=4\alpha+40^{\circ}$$

D)$$\beta=5\alpha-20^{\circ}$$

I solved it by taking $$\cos x=1-2\sin^232^{\circ}$$

Therefore $$x=64$$

So the two values of $$x$$ can be $$64$$ and $$334$$, so B) should be the right answer, but the answer is actually c. Where am I going wrong?

Note that: $$\cos x=\cos 64^\circ \Rightarrow x=\pm64^\circ+ 360^\circ n \\ n=0 \Rightarrow x=64^\circ\\ n=1 \Rightarrow x=-64^\circ +360^\circ =296^\circ$$
$$360-64=296=\beta$$
$$4\alpha=4\cdot64=256$$
Since you found that $$x=64$$, you could draw the reference triangle for $$\cos(64^\circ)$$
Taking $$360-64=296$$ would produce the only angle for $$0\leq x \leq 2\pi$$ which would have the same cosine value. The reference triangle for $$\cos(296^\circ)$$ is
If you wanted to compute the values of $$\cos(64^\circ)$$ and $$\cos(296^\circ)$$ then you would find that they are both $$\sin\big(\frac{13\pi}{90}\big)$$.