the following equation involves a trigonometric equation in quadratic form\n2 sin²x = 5 sinx + 7\nselect the…

Explanation:

Step1: Rewrite the equation

$$2\sin^{2}x - 5\sin x - 7 = 0$$ Let (t=\sin x), then the equation becomes (2t^{2}-5t - 7 = 0)

Step2: Solve the quadratic equation

Use the quadratic formula (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (at^{2}+bt + c = 0). Here (a = 2), (b=-5), (c=-7) [ \begin{align*} t&=\frac{5\pm\sqrt{(-5)^{2}-4\times2\times(-7)}}{2\times2}\ &=\frac{5\pm\sqrt{25 + 56}}{4}\ &=\frac{5\pm\sqrt{81}}{4}\ &=\frac{5\pm9}{4} \end{align*} ] We get (t_{1}=\frac{5 + 9}{4}=\frac{14}{4}=\frac{7}{2}) and (t_{2}=\frac{5-9}{4}=-1)

Step3: Substitute back (t = \sin x)

Since (-1\leqslant\sin x\leqslant1), for (t=\frac{7}{2}), (\sin x=\frac{7}{2}) has no solution. For (t=-1), (\sin x=-1), then (x=\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z})

Answer:

A. (x=\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z})