2007_Question1c

Since \sin^2+\cos^2=1, we can rewrite the equation as 2(1-\alpha^2)+7\alpha-5=0, where \alpha=\sin x. This is equivalent to 2\alpha^2-7\alpha+3=0, which factors as (2\alpha-1)(\alpha-3)=0, and so has solutions \alpha=\frac12 and \alpha=3. The latter is impossible, since \vert\sin x\vert\leqslant1; the former has two solutions in the given range (x=\frac{\pi}6 and x=\frac{5\pi}6), so the answer is (b).

Correct!!!