0

Guest

Courses New

Find the no. of solution of given trigonometric equation where 0≤x≤5π(3+cosx)^2 = 4- 2sin^8x

Find the no. of solution of given trigonometric equation where 0≤x≤5π(3+cosx)^2 = 4- 2sin^8x 

Question Image
Grade:12th pass

2 Answers

Susmita
425 Points
5 years ago
It is very easy to see that this equation will hold only for x=pi,x=3pi,x=5pi.
cospi=cos3pi=cos5pi=-1 and sin term will be zero.
(3-1)2=4
You can put these values and check.
So number of solution is 3.
Aditya Gupta
2081 Points
5 years ago
Let cosx=c and sinx be s
LHS is (3+c)^2, whose minimum value is 4
RHS is 4-2s^8 whose maximum value is 4.
So for them to be equal, both must equal 4.
So s=0 and c=-1
As c=-1 automatically implies s=0, hence we only need to focus on the solutions of cosx=-1=cosπ 
x= odd mutiple of pi
So x can be pi, 3pi, 5pi.
So there are exactly three solutions.

Think You Can Provide A Better Answer ?

Other Related Questions on Trigonometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More