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the number of values of x satisfying the equation sinx +sin5x=sin3x such that x€[0,pi]
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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
Community Answer
the number of values of x satisfying the equation sinx +sin5x=sin3x su...
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the number of values of x satisfying the equation sinx +sin5x=sin3x such that x€[0,pi]
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the number of values of x satisfying the equation sinx +sin5x=sin3x such that x€[0,pi] for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about the number of values of x satisfying the equation sinx +sin5x=sin3x such that x€[0,pi] covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for the number of values of x satisfying the equation sinx +sin5x=sin3x such that x€[0,pi].
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