Question

the number of solution of sin 3x = cos 2x in interval (π/2,π)

Answer 1

Answer:

1, x = 9π/10.

Step-by-step explanation:

Hi,

Given sin3x = cos2x

=> sin3x = sin(π/2 - 2x)

We now that sinA = sinB

=> A = nπ + (-)ⁿB, where n is an integer

Using the above identity, we get

3x = nπ + (-1)ⁿ(π/2-2x)

If n= 1 , x = π- π/2=> x = π/2∉ (π/2,π)

If n= 2 , x = π/2∉ (π/2,π)

If n=3, x = 5π/2 ∉ (π/2,π)

If n= 4 , x = 9π/10∈ (π/2,π)

If n=5, x = 9π/2 = π∉ (π/2,π)

Now, for all negative integers x would be negative.

For all value of n > 5, solution > π

Hence the only possible solution is for n = 4

and x = 9π/10.

Hope, it helped !

Answer 2

Answer:

Thanks for answering This question