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Question

How many values of x in the interval [0,2π] satisfies the equation sin6x=1+cos43x


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Solution

Maximum value of sin6 x is 1 and minimum value of 1+ cos4 3x is 1.

sin6 x 1 and sin6x = 1 + cos4 3x 1

1 sin6 x 1

sin6 x = 1

cos4 3x = sin6 x - 1 = 1 - 1 = 0

sin6 x = 1 and cos4 3x = 0

x=(2n+1)π2 and 3x=(2m+1)π2.
x=(2n+1)π2 and x=(2m+1)π6.
We want to find the common values of (2n+1)π2and(2m+1)π6. For some value of m and n,

both these expressions will be equal corresponding to common terms.
(2n+1)π2=(2m+1)π6

2m+1=6n+3

2m=6n+2
m = 3n + 1

This means for any value of n, there will be a corresponding m.

All the values of (2n + 1) π2 is also part of the sequence (2m + 1) π6.

So the solution is x = (2n + 1) π2.

The values in the interval [0, 2π] are π2and3π2 corresponding to n = 0 and n = 1.

So the number of solutions in the interval [0, 2π] is 2.


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