the number of values of x satisfying the equation sinx +sin5x=sin3x su...
Number of Values of x Satisfying the Equation sinx sin5x = sin3x
The given equation is:
sinx sin5x = sin3x
Let's simplify this equation:
sinx (sin2x.sin3x) = sin3x
sinx (2sinx.cosx.3sinx-4sin3x) = 0
sinx(6sin^2x-4sin^3x-2cosx) = 0
sinx(2sin^2x-4sin^3x-cosx) = 0
Case 1: sinx = 0
This means that x = nπ where n is an integer.
There are a total of (π+1) values of x where sinx = 0 in the interval [0, π].
Case 2: 2sin^2x-4sin^3x-cosx = 0
Let's define a function f(x) = 2sin^2x-4sin^3x-cosx
f(x) is continuous in the interval [0, π].
f(0) = -1 and f(π) = 1.
Therefore, by the intermediate value theorem, there exists at least one value of x in the interval [0, π] where f(x) = 0.
Let's find the derivative of f(x):
f'(x) = 4sinx(3sinx-2sin^2x+cosx)
f'(x) = 4sinx(3sinx-2(1-cos^2x)+cosx)
f'(x) = 4sinx(3sinx-2+2cos^2x+cosx)
f'(x) = 4sinx(3sinx+2cos^2x+cosx-2)
f'(x) = 4sinx((3sinx+1)(sinx+2)-4)
f'(x) = 4sinx((3sinx+1)(sinx+2)-4)
Therefore, f'(x) = 0 at x = 0, π/3, and π.
f''(x) = 4(3sinx+1)(cosx)+8sinx(2sinx+cosx-3)
At x = 0, f''(0) = -4 and f''(π) = 4.
Therefore, f(x) has a local maximum at x = 0 and a local minimum at x = π.
By analyzing the sign of f'(x) and f''(x) in the different intervals, we can conclude that f(x) has:
- Two roots in the interval [0, π