Question

please refer the attachment and solve it ​

Answer 1

Required Answer:-

Given:

$\rm \mapsto \sin(x) + { \sin}^{2}(x) + { \sin}^{3}(x) = 1$

To find:

$\rm \cos^{6} (x) - 4 \cos^{4} (x) + 8 { \cos}^{2} (x) = ?$

Solution:

Given,

$\rm \sin(x) + { \sin}^{2}(x) + { \sin}^{3}(x) = 1$

$\rm \implies \sin(x) + { \sin}^{3}(x) = 1 - { \sin}^{2}(x)$

$\rm \implies \sin(x) \{1 + { \sin}^{2}(x) \} = 1 - { \sin}^{2}(x)$

We know that,

$\rm { \sin}^{2}(x) + { \cos}^{2} (x) = 1$

$\rm \implies { \sin}^{2}(x) = 1 - { \cos}^{2} (x)$

$\rm \implies { \cos}^{2} (x) = 1 - { \sin}^{2}(x)$

Therefore,

$\rm \implies \sin(x) \{1 + { \sin}^{2}(x) \} = { \cos}^{2} (x)$

$\rm \implies \sin(x) \{1 + 1 - { \cos}^{2}(x) \} = { \cos}^{2} (x)$

$\rm \implies \sin(x) \{2- { \cos}^{2}(x) \} = { \cos}^{2} (x)$

On squaring both sides, we get,

$\rm \implies \sin^{2}(x) \{2- { \cos}^{2}(x) \}^{2} = { \cos}^{4} (x)$

$\rm \implies(1 - \cos^{2}(x) )\{2- { \cos}^{2}(x) \}^{2} = { \cos}^{4} (x)$

$\rm \implies(1 - \cos^{2}(x) )\{ {2}^{2} + { \cos}^{4}(x) - 4 \cos ^{4} (x) \} = { \cos}^{4} (x)$

$\rm \implies(1 - \cos^{2}(x) )\{4 + { \cos}^{4}(x) - 4 \cos ^{2} (x) \} = { \cos}^{4} (x)$

$\rm \implies 4 + { \cos}^{4}(x) - 4 { \cos}^{2}(x) - 4 { \cos}^{2} (x) - { \cos }^{6} (x) + 4 { \cos }^{4} (x) = { \cos}^{4}(x)$

$\rm \implies 4 + - 8{ \cos}^{2} (x) - { \cos }^{6} (x) + 4 { \cos }^{4} (x) = 0$

$\rm \implies 8{ \cos}^{2} (x) + { \cos }^{6} (x) - 4 { \cos }^{4} (x) = 4$

$\rm \implies { \cos }^{6} (x) - 4 { \cos }^{4} (x) + 8 { \cos}^{2}(x) = 4$

Hence, Required Answer is 4.

Answer:

$\rm { \cos }^{6} (x) - 4 { \cos }^{4} (x) + 8 { \cos}^{2}(x) = 4$

Formula Used:

$\rm \mapsto { \sin}^{2}(x) + { \cos}^{2} (x) = 1$

Answer 2

Step-by-step explanation:

ANSWER IS 4...................