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Find the no. of solution of given trigonometric equation where 0≤x≤5π(3+cosx)^2 = 4- 2sin^8x
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=4You can put these values and check.So number of solution is 3.
Let cosx=c and sinx be sLHS is (3+c)^2, whose minimum value is 4RHS is 4-2s^8 whose maximum value is 4.So for them to be equal, both must equal 4.So s=0 and c=-1As c=-1 automatically implies s=0, hence we only need to focus on the solutions of cosx=-1=cosπ x= odd mutiple of piSo x can be pi, 3pi, 5pi.So there are exactly three solutions.
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