Question

Find general solution for given expression...

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Answer 1

Answer:

$\large \bold\red{x = n\pi \pm {( - 1)}^{n} \frac{\pi}{4}\:\:;n=Integer }$

Step-by-step explanation:

Given,

$3 { \sin }^{4} x + { \cos}^{4} x = 1$

But,

We know that,

• ${ \cos}^{2} \alpha = 1 - { \sin}^{2} \alpha$

Therefore,

Substituting the values,

We get,

$= > 3 { \sin }^{4} x + {(1 - { \sin }^{2}x) }^{2} = 1 \\ \\ = > 3 { \sin }^{4} x + \cancel{ 1} - 2 { \sin}^{2} x + { \sin}^{4} x = \cancel{1} \\ \\ = > 4 { \sin }^{4} x - 2 { \sin }^{2} x = 0 \\ \\ = > 2 { \sin }^{2} x(2 { \sin}^{2} x - 1) = 0$

Therefore,

We have two Equations,

$\bold\purple{\underline{Case\:I}}$

$= > 2 { \sin }^{2} x = 0 \\ \\ = > { \sin }^{2} x = 0 \\ \\ = > \sin x= 0$

Therefore,

General Solution will be,

$\large \boxed{ \bold {x = n\pi \: }}$

• Where, n is an integer

$\bold\purple{\underline{Case\:II}}$

$= > 2{ \sin }^{2} x - 1 = 0 \\ \\ = >2 { \sin }^{2} x = 1 \\ \\ = > { \sin }^{2} x = \frac{1}{2} \\ \\ = > \sin x = \pm \frac{1}{ \sqrt{2} } \\ \\ = > \sin x = \sin( \pm \frac{\pi}{4} )$

Therefore,

General Solution will be,

$\large \boxed{ \bold{x = n\pi \pm {( - 1)}^{n} \frac{\pi}{4} }}$

• Where, n is an integer

Hence,

Combining both the results,

We have general solution for the given Equation as,

$\large \boxed{ \large\bold \pink{x = n\pi \pm {( - 1)}^{n} \frac{\pi}{4} }}$

• Where , n is an integer.