Question

If 4^2sin^2 x . 16^ tan^2 x. 2^4cos^2 x = 256 such that 0 < x < π/2

 then x is equal
a. π/3
b. π/6
c. π/4
d. π/24

Answer 1

Answer:

C) $x \: = \frac{\pi}{4}$

Step-by-step explanation:

Given---->

${4}^{2 { \sin }^{2} x \:} \: {16}^{ { \tan }^{2}x } \: {2}^{4 { \cos }^{2}x } \: = \: 256 \\ and \: 0 \leqslant \: x \: \leqslant \frac{\pi}{2}$

To find ---->

$value \: of \: x$

Concept used ---->

1)

$( \: { {a}^{m} })^{n } = \: {a}^{mn}$

2)

${a}^{m} \: {a}^{n} \: = \: {a}^{mn}$

3)

${ \sin }^{2} x \: + \: { \cos }^{2} x \: = 1$

4)

$1 + { \tan }^{2} x \: = { \sec }^{2}x$

Solution-----> ATQ,

=> ${4}^{2 { \sin }^{2}x \: } {16}^{ { \tan }^{2}x } \: {2}^{4 { \cos }^{2} x} \: = 256$

=> $( \: { {2}^{2}) }^{2 { \sin }^{2}x } \: ( \: { {2}^{4}) }^{ { { \tan }^{2} }x } \: {2}^{4 { \cos }^{2}x } \: = {2}^{8}$

=> ${2}^{4 { \sin }^{2}x } \: {2}^{4 { \tan }^{2}x } \: {2}^{4 { \cos }^{2}x } \: = {2}^{8}$

=> ${2}^{4 { \sin }^{2}x \: + 4 { \tan }^{2}x \: + 4 { \cos }^{2}x } \: = {2}^{8}$

$comparing \: exponent \: from \: both \: sides \: we \: get$

=> $4 { \sin }^{2}x \: + 4 { \tan }^{2}x \: + 4 { \cos }^{2}x \: = \: 8$

$taking \: 4 \: common$

=> $4 \: ( \: { \sin }^{2}x \: + { \cos }^{2}x \: + { \tan }^{2}x \:) = 8$

=> $1 \: + { \tan }^{2}x \: = 2$

=> ${ \sec }^{2}x \: = 2$

=> ${ \cos }^{2}x \: = \frac{1}{2}$

$if \: \cos(x) \: = - \frac{1 \ }{ \sqrt{2} } \: impossible \: because \: 0 \leqslant x \leqslant \frac{\pi}{2}$

$if \: \cos(x) = \frac{1}{ \sqrt{2} }$

=> $\cos(x) = \cos( \frac{\pi}{4} )$

=> $x \: = \frac{\pi}{4}$