Question

2cos^2x+3sinx=0 solve the equation

Answer 1

Answer:

The values of $\sin x = 2$ or $\sin x = - 1 / 2$

To find:

The value of sin x

Solution:

Given :

$2 \cos ^ { 2 } x + 3 \sin x = 0$

As

$\begin{array} { c } { \sin ^ { 2 } x + \cos ^ { 2 } x = 1 } \\\\ { \cos ^ { 2 } x = 1 - \sin ^ { 2 } x } \end{array}$

Replacing $\cos ^ { 2 } x = 1 - \sin ^ { 2 } x$ in the given equation,

$\begin{array} { l } { 2 \left( 1 - \sin ^ { 2 } \mathrm { x } \right) + 3 \sin \mathrm { x } = 0 } \\\\ { 2 - 2 \sin ^ { 2 } \mathrm { x } + 3 \sin \mathrm { x } = 0 } \\\\ { 2 \sin ^ { 2 } \mathrm { x } - 3 \sin \mathrm { x } - 2 = 0 } \\\\ { 2 \sin ^ { 2 } \mathrm { x } - 4 \sin \mathrm { x } + \sin \mathrm { x } - 2 = 0 } \\\\ { 2 \sin \mathrm { x } ( \sin \mathrm { x } - 2 ) + 1 ( \sin \mathrm { x } - 2 ) = 0 } \\\\ { ( \sin \mathrm { x } - 2 ) ( 2 \sin \mathrm { x } + 1 ) = 0 } \end{array}$

Either

$\begin{array} { l } { \sin x - 2 = 0 } \\\\ { \sin x = 2 } \end{array}$

or

$\begin{array} { c } { 2 \sin x + 1 = 0 } \\\\ { \sin x = - \frac { 1 } { 2 } } \end{array}$

The values of $\sin x = 2$ or $\sin x = - 1 / 2$