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Grade: 12th pass
        
Find the no. of solution of given trigonometric equation where 0≤x≤5π(3+cosx)^2 = 4- 2sin^8x 
one year ago

Answers : (2)

Susmita
425 Points
							
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.
one year ago
Aditya Gupta
1672 Points
							
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.
11 months ago
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