Question

Solve the trigonometric equation ${{\sin }^{4}}x+{{\cos }^{4}}x=\dfrac{5}{8}$ for the value of x.

Answer 1

Firstly we will write ${{\sin }^{4}}x$ as ${{({{\sin }^{2}}x)}^{2}}$ and $co{{s}^{4}}x$ as ${{(co{{s}^{2}}x)}^{2}}$ and obtain expression${{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}=\dfrac{5}{8}$. After that we will add sin2x on both sides and put $\sin 2x=2\sin x\times \text{cosx}$, which will make a perfect square on the left hand side, which will help us in reducing the complex equation question to a simpler form of equation. By substituting this value we will continue and get the required value of x.

Complete step-by-step answer:
We know that trigonometry is a branch of mathematics that studies relationships between the side lengths and the angles of triangles.
Here, we are given that ${{\sin }^{4}}x+{{\cos }^{4}}x=\dfrac{5}{8}$.
We can write this given equation as :
${{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}=\dfrac{5}{8}...........\left( 1 \right)$
On adding $2{{\sin }^{2}}x{{\cos }^{2}}x$ on both sides of the equation (1), we get:
${{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}+2{{\sin }^{2}}x{{\cos }^{2}}x=\dfrac{5}{8}+2{{\sin }^{2}}x{{\cos }^{2}}x$
The Left hand side of the equation can be written in the form of a perfect square by using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ as:
${{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}=\dfrac{5}{8}+2{{\sin }^{2}}x{{\cos }^{2}}x$
Since, we have the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, using this in above equation , we get:
$1-2{{\sin }^{2}}x{{\cos }^{2}}x=\dfrac{5}{8}$
Now, on multiplying both sides of the equation by 2, we get:
$\begin{aligned} & 2-4{{\sin }^{2}}x{{\cos }^{2}}x=\dfrac{5}{4} \\\ & \Rightarrow 2-{{\left( 2\sin x\cos x \right)}^{2}}=\dfrac{5}{4} \\\ \end{aligned}$
On using the trigonometric identity that $\sin 2x=2\sin x\cos x$, we get:
$\begin{aligned} & 2-{{\left( \sin 2x \right)}^{2}}=\dfrac{5}{4} \\\ & \Rightarrow {{\sin }^{2}}2x=2-\dfrac{5}{4}=\dfrac{3}{4} \\\ \end{aligned}$
Now, taking square root on both sides of the equation, we get:
$\sin 2x=\pm \dfrac{\sqrt{3}}{2}$
We know that the value $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$, so using this value, we get:
$\sin 2x=\pm \sin \left( \dfrac{\pi }{3} \right)$
We know that sin2x will repeat its value after an interval of$\dfrac{\pi }{2}$ that is ${{90}^{0}}$. Therefore, we have:
$2x=n\pi \pm \dfrac{\pi }{3}$
$\Rightarrow x=\dfrac{n\pi }{2}\pm \dfrac{\pi }{6}$ , here ‘n’ is an integer greater than or equal to 0.
Hence, the solution of the given equation is $x=\dfrac{n\pi }{2}\pm \dfrac{\pi }{6}$.

Note: Here, students should note that the angle is 2x, so we have to divide the obtained solution by 2 to get the value of x. Also students should know the way of forming perfect squares from a given sum of two squares as in the case of ${{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}$. Also, one must know the trigonometric values of different angles such as $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$ and also that period of function sin2x is $\dfrac{\pi }{2}$. Try not to make any calculation mistakes as each step needs to be correct to get the correct solution.