Question

solve this .........☺️

Answer 1

Q : 7 If $x \in [\frac{-5\pi}{2}, \frac{5\pi}{2}]$ ,the greatest positive solution of
$1 + \sin^4(x) = \cos^2(3x)$ .

Ans: 7 : We have,

Steps:
$1) LHS = 1 + sin^4(x) \\ => LHS_{min}= 1+0= 1 \\ RHS= \cos^2(3x) \\ RHS_{max}=1$

2)This is possible when
LHS = RHS = 1
$1 + \sin^4(x) = 1 \\ => \sin^4(x) = 0 \\ => \sin(x)= 0\\ => x = n\pi : [n\in I] \\ =>x =± \pi, ±2\pi$
=> Maximum positive value of x in given x = 2π

Q: 8 : If
$\cos(x) = \sqrt{1-\sin(2x)}, 0<x < \pi$ ,
then a value of x is :
Ans: 8 :
1# We have,
$\cos(x) = \sqrt{1-\sin(2x)},$
=> cos(x) is positive.
=> 0 <x <π/2

2# : $\cos^2(x) = 1 - \sin(2x) \\ => \cos^2(x) = \sin^2(x) + \\ \cos^2(x) -2 \sin(x) \cos(x) \\ => \sin^2(x) = 2\sin(x) cos(x) \\ => \sin(x) = 2\cos(x) : as[ \sin(x) \neq 0] => \tan(x) = 2 \\ => x = \tan^{-1}(2)$