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=(1+3)/(4),y=(1-3)/(4) \\ y=1,y=-(1)/(2) \end{gathered}`

Note that we substituted sinx equals y.

The result means;

`\sin x=1,\sin x=-(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=(\pi)/(2)+2\pi n \\ \text{When,} \\ \sin x=-(1)/(2),\text{ then} \\ x=(7\pi)/(6)+2\pi n,x=(11\pi)/(6)+2\pi n \end{gathered}`

ANSWER:

`\begin{gathered} B\colon\text{False} \\ (5\pi)/(2)\text{ is not a solution} \end{gathered}`