Question

How to solve this inequation? ${\sin ^4}x + {\cos ^4}x \geqslant \dfrac{1}{2}$

Answer 1

Hint: Use trigonometric identity related to sine and cosine function and try to simplify the expression ${\sin ^4}x + {\cos ^4}x$ in less power. Then find the solution for the given condition, first find the principal solution and then try to find the general solution.

Complete step by step answer:
 In order to solve the given inequation ${\sin ^4}x + {\cos ^4}x \geqslant \dfrac{1}{2}$
We will try to simplify the trigonometric expression ${\sin ^4}x + {\cos ^4}x$ first,
${\sin ^4}x + {\cos ^4}x$
We can write it as
${\left( {{{\sin }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^2}$
From the algebraic formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$, we can write the above trigonometric expression as
\[
{\left( {{{\sin }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^2} \\
\Rightarrow{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\;\;\left[ {\because {a^2} + {b^2} = {{\left( {a + b} \right)}^2} - 2ab} \right] \\ \]
Now, from the trigonometric identity of sine and cosine function, i.e. ${\sin ^2}x + {\cos ^2}x = 1$
We can further simplify the trigonometric expression as
\[
{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\; \\
\Rightarrow{\left( 1 \right)^2} - 2{\sin ^2}x{\cos ^2}x\; \\
\Rightarrow 1 - 2{\sin ^2}x{\cos ^2}x\; \\ \]
From the trigonometric identity of compound sine angle formula, we know that
$\sin 2x = 2\sin x\cos x$, from this we can write the trigonometric expression further as
\[
1 - 2{\sin ^2}x{\cos ^2}x\; \\
\Rightarrow 1 - (2\sin x\cos x) \times (2\sin x\cos x) \times \dfrac{1}{2} \\
\Rightarrow 1 - \sin 2x \times \sin 2x \times \dfrac{1}{2} \\
\Rightarrow 1 - \dfrac{{{{\sin }^2}2x}}{2} \\ \]
So we have simplified the trigonometric expression ${\sin ^4}x + {\cos ^4}x$ to \[1 - \dfrac{{{{\sin }^2}2x}}{2}\]. Now from the inequation, we can write
$
{\sin ^4}x + {\cos ^4}x \geqslant \dfrac{1}{2} \\
\Rightarrow 1 - \dfrac{{{{\sin }^2}2x}}{2} \geqslant \dfrac{1}{2} \\
\Rightarrow 1 - \dfrac{1}{2} \geqslant \dfrac{{{{\sin }^2}2x}}{2} \\
\Rightarrow \dfrac{1}{2} \geqslant \dfrac{{{{\sin }^2}2x}}{2} \\
\Rightarrow 1 \geqslant {\sin ^2}2x \\
\Rightarrow \sin 2x \leqslant \pm 1 \\ $
We have to solve this in two cases, first for positive one and second for negative one
Taking the first case,
$ \Rightarrow \sin 2x \leqslant 1$
Since this is an identity as the range of sine function lies between $ - 1\;{\text{to}}\;1$
So this answer to this case is $2x \in R \Rightarrow x \in R$ as it is an identity
Solving the second case,
$ \Rightarrow \sin 2x \leqslant - 1$
This case is not possible because range of sine function is always greater than $ - 1$

Therefore the inequation ${\sin ^4}x + {\cos ^4}x \geqslant \dfrac{1}{2}$ hold true for all values of $x\;{\text{i}}{\text{.e}}{\text{.}}\;x \in R$

Note: In order to find the solutions for trigonometric inequalities, you have to check for the domain, range, and also where the given trigonometric function is not defined, so that you can exclude that from your solution.