Question

The value 57 is a solution for the equation 2 sin? x - sin x-1 = 0). O A. True O B. False

Answer 1

To solve for

$2\sin ^2x-\sin x-1=0$

We shall first substitute for sinx = y.

Therefore, we can re-write this as;

$\begin{gathered} 2y^2-y-1=0 \\ We\text{ shall solve using the quadratic equation formula as follows;} \\ y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=2,b=-1,c=-1 \\ y=\frac{-(-1)\pm\sqrt[]{(-1)^2-4(2)(-1)}_{}}{2(2)} \\ y=\frac{1\pm\sqrt[]{1+8}}{4} \\ y=\frac{1\pm\sqrt[]{9}}{4} \\ y=\frac{1+3}{4},y=\frac{1-3}{4} \\ y=1,y=-\frac{1}{2} \end{gathered}$

Note that we substituted sinx equals y.

The result means;

$\sin x=1,\sin x=-\frac{1}{2}$

Taking values of sinx on the unit circle, the general solutions to sinx are;

$\begin{gathered} \text{When,} \\ \sin x=1,\text{ then} \\ x=\frac{\pi}{2}+2\pi n \\ \text{When,} \\ \sin x=-\frac{1}{2},\text{ then} \\ x=\frac{7\pi}{6}+2\pi n,x=\frac{11\pi}{6}+2\pi n \end{gathered}$

ANSWER:

$\begin{gathered} B\colon\text{False} \\ \frac{5\pi}{2}\text{ is not a solution} \end{gathered}$