Question

The number of real solutions of the equation ${\cos ^5}x + {\sin ^3}x = 1$ in the interval $[0,2\pi ]$ is
$A) 2$
$B) 1$
$C) 3$
$D) Infinite$

Answer 1

In this problem we need to find the solution of the given equations with the given intervals. For that we need to solve the trigonometric values of the equation with the class interval. In this process we have to explain the particular change in the expression. We need to find the number of solutions of the given equation that will take place in two cases and with formula required. By basic mathematical calculation and complete step by step explanation we can solve this.

Formula used: ${\cos ^{ - 1}}(0) = \dfrac{\pi }{2}$
${\sin ^{ - 1}}(0) = 0,2\pi$
${\cos ^{ - 1}}(1) = 0,2\pi$
${\sin ^{ - 1}}(1) = \dfrac{\pi }{2}$

Complete step-by-step solution:
Let us consider the given equation: ${\cos ^5}x + {\sin ^3}x = 1$
In this equation there are two cases, they are
$\Rightarrow {\cos ^5}x + {\sin ^3}x = 1$
$\Rightarrow 0 + 1 = 1$
$\Rightarrow 1 + 0 = 1$
Case (i) $0 + 1 = 1$
$\Rightarrow {\cos ^5}x + {\sin ^3}x = 1$
Now, we can rearrange the equation
$\Rightarrow {(\cos x)^5} + {(\sin x)^3} = 1$
Now compare the equation with case (i), we get
$\Rightarrow {(\cos x)^5} = 0,{(\sin x)^3} = 1$
$\Rightarrow \cos x = 0,\sin x = 1$
Now separate the value of x, we get
$\Rightarrow x = {\cos ^{ - 1}}0,x = {\sin ^{ - 1}}1$
By using formula mentioned in formula used, we get
$\Rightarrow x = 0,2\pi ,\dfrac{\pi }{2}$
Case (ii) $1 + 0 = 1$
$\Rightarrow {\cos ^5}x + {\sin ^3}x = 1$
Now, we can rearrange the equation
$\Rightarrow {(\cos x)^5} + {(\sin x)^3} = 1$
Now compare the equation with case (i), we get
$\Rightarrow {(\cos x)^5} = 1,{(\sin x)^3} = 0$
$\Rightarrow \cos x = 1,\sin x = 0$
Now separate the value of x, we get
$\Rightarrow x = {\cos ^{ - 1}}1,x = {\sin ^{ - 1}}0$
By using formula mentioned in formula used, we get
$\Rightarrow x = 0,2\pi$
Therefore the real solutions of equations are $x = 0,\dfrac{\pi }{2},2\pi$
The number of real solutions is $3$.

Option C is the correct answer.

Note: This problem needs attention on trigonometric equations solvable with class intervals. Here we asked to find the number of real solutions for which we need to find the solution in two cases. We should be familiar with trigonometric identities with which we have found the solution. This kind problem is based on solving class intervals in the given equation which can be either in any form. We have to be familiar with that so that we can solve it.