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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?

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Note that: $$\cos x=\cos 64^\circ \Rightarrow x=\pm64^\circ+ 360^\circ n \\ n=0 \Rightarrow x=64^\circ\\ n=1 \Rightarrow x=-64^\circ +360^\circ =296^\circ $$

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$$360-64=296=\beta$$

$$4\alpha=4\cdot64=256$$

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Since you found that $x=64$, you could draw the reference triangle for $\cos(64^\circ)$

enter image description here

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

enter image description here

If you wanted to compute the values of $\cos(64^\circ)$ and $\cos(296^\circ)$ then you would find that they are both $\sin\big(\frac{13\pi}{90}\big)$.

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