Question

2cos²x=7sin x-2 then x=?

Answer 1

Answer---->

$principal \: value \: of \: x$

$x = \dfrac{\pi}{6} and \: \dfrac{\ \: 5\pi}{6}$

$general \: value \: of \: x$

$x = n\pi + { (- 1)}^{n} \dfrac{\pi}{6}$

Step-by-step explanation:

Given---->

$2 { \cos }^{2} x = 7 \: \sin \: x - 2$

To find ---->

$value \: of \: x$

Concept used ---->

1)

$1 - { \sin }^{2}x = { \cos }^{2}x$

2)

$general \: value \: of \: \sin$

$n\pi + { (- 1)}^{n} \alpha$

3)

$value \: of \: sinx \: is \: greater \: than \: equal \: to \: ( - 1) \: and \: less \: than \: equal \: to \: 1$

$- 1 \leqslant \sin \: x \leqslant 1$

Solution----> ATQ,

$2 \cos^{2}x = 7 \sin \: x \: - 2$

=> $2 \ \: ( \: 1 - { \sin \: }^{2}x \: ) = 7 \sin \: x \: - 2$

=> $2 - 2 \: { \sin }^{2}x \: - \: 7 \sin \: x + 2 = 0$

=> $- 2 \sin ^{2}x \: - 7 \sin \: x + 4 \: = 0$

=> $2 \sin^{2}x \: + 7 \sin \: x \: - 4 = 0$

=> $2 \sin^{2}x + 8 \sin \: x - \sin \: x - 4 = 0$

=> $2 \sin \: x( \: \sin \: x \: + 4 \: ) - 1( \: \sin \: x \: + 4 \: ) = 0$

=> $( \: \sin \: x \: + 4 \: ) \: ( \: 2 \sin \: x \: - 1 \: ) = 0$

$if$

$\sin \: x \: + 4 = 0$

=> $\sin \: x \: = - 4 \: \: ( \: impossible \: )$

$because \: \sin \: x \: can \: not \: be \: less \: than \: ( \: - 1 \: ) \: so \: it \: is \: impossible$

$if$

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

=> $2 \sin \: x \: = 1$

=> $\sin \: x \: = \frac{1}{2}$

$principal \: value \: of \: x \: is$

$x \: = \dfrac{\pi}{6} and \: \dfrac{5\pi}{6}$

$general \: value \: of \: x$

$\sin \: x \: = \frac{1}{2}$

=> $\sin \: x \: = \frac{\pi}{6}$

=> $sinx = sin \: ( \: n\pi \: + ( - 1) ^{n} \: \frac{\pi}{6} \: )$

=> $x = n\pi \: + ( - 1 \: ) ^{n} \frac{\pi}{6}$