The number of solutions of the equation $\cos 2\theta \cos \frac{\theta}{2} + \cos \frac{5\theta}{2} = 2\cos^3 \frac{5\theta}{2}$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$is :
Solving the Trigonometric Equation in the Interval
We need to find the number of solutions for the equation $ \cos 2\theta \cos \frac{\theta}{2} + \cos \frac{5\theta}{2} = 2\cos^3 \frac{5\theta}{2} $ in the interval $ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $.
Simplify the Equation
Use the product-to-sum identity $ 2 \cos A \cos B = \cos(A+B) + \cos(A-B) $. Applying this to the first term:
$ \cos 2\theta \cos \frac{\theta}{2} = \frac{1}{2} \left( \cos\left(2\theta + \frac{\theta}{2}\right) + \cos\left(2\theta - \frac{\theta}{2}\right) \right) = \frac{1}{2} \left( \cos \frac{5\theta}{2} + \cos \frac{3\theta}{2} \right) $Substitute this back into the original equation:
$ \frac{1}{2} \left( \cos \frac{5\theta}{2} + \cos \frac{3\theta}{2} \right) + \cos \frac{5\theta}{2} = 2\cos^3 \frac{5\theta}{2} $Combine terms:
$ \frac{3}{2} \cos \frac{5\theta}{2} + \frac{1}{2} \cos \frac{3\theta}{2} = 2\cos^3 \frac{5\theta}{2} $Use the identity $ 2\cos^3 x = \frac{1}{2}(3\cos x + \cos 3x) $ with $ x = \frac{5\theta}{2} $. The right side becomes:
$ 2\cos^3 \frac{5\theta}{2} = \frac{1}{2} \left( 3\cos \frac{5\theta}{2} + \cos \frac{15\theta}{2} \right) $Substituting this into the combined equation:
$ \frac{3}{2} \cos \frac{5\theta}{2} + \frac{1}{2} \cos \frac{3\theta}{2} = \frac{3}{2} \cos \frac{5\theta}{2} + \frac{1}{2} \cos \frac{15\theta}{2} $Simplify by cancelling terms:
$ \cos \frac{3\theta}{2} = \cos \frac{15\theta}{2} $Finding General Solutions
The equation $ \cos A = \cos B $ implies $ A = 2n\pi \pm B $, where $ n $ is an integer.
- Case 1: $ \frac{15\theta}{2} = 2n\pi + \frac{3\theta}{2} $ $ \frac{15\theta}{2} - \frac{3\theta}{2} = 2n\pi $ $ \frac{12\theta}{2} = 2n\pi $ $ 6\theta = 2n\pi $ $ \theta = \frac{n\pi}{3} $
- Case 2: $ \frac{15\theta}{2} = 2n\pi - \frac{3\theta}{2} $ $ \frac{15\theta}{2} + \frac{3\theta}{2} = 2n\pi $ $ \frac{18\theta}{2} = 2n\pi $ $ 9\theta = 2n\pi $ $ \theta = \frac{2n\pi}{9} $
Counting Solutions in the Interval $ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $
We need to find the integer values of $ n $ for each case such that $ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} $.
- For $ \theta = \frac{n\pi}{3} $: $ -\frac{\pi}{2} \le \frac{n\pi}{3} \le \frac{\pi}{2} $ $ -\frac{1}{2} \le \frac{n}{3} \le \frac{1}{2} $ $ -\frac{3}{2} \le n \le \frac{3}{2} $ Integer values for $ n $: $ -1, 0, 1 $. Solutions: $ \theta = -\frac{\pi}{3}, 0, \frac{\pi}{3} $. (3 distinct solutions)
- For $ \theta = \frac{2n\pi}{9} $: $ -\frac{\pi}{2} \le \frac{2n\pi}{9} \le \frac{\pi}{2} $ $ -\frac{1}{2} \le \frac{2n}{9} \le \frac{1}{2} $ $ -\frac{9}{2} \le 2n \le \frac{9}{2} $ $ -2.25 \le n \le 2.25 $ Integer values for $ n $: $ -2, -1, 0, 1, 2 $. Solutions: $ \theta = -\frac{4\pi}{9}, -\frac{2\pi}{9}, 0, \frac{2\pi}{9}, \frac{4\pi}{9} $. (5 distinct solutions)
Total Distinct Solutions
Combine the solutions from both cases and count the distinct ones.
- Case 1 solutions: $ \{-\frac{\pi}{3}, 0, \frac{\pi}{3}\} $
- Case 2 solutions: $ \{-\frac{4\pi}{9}, -\frac{2\pi}{9}, 0, \frac{2\pi}{9}, \frac{4\pi}{9}\} $
The solution $ \theta = 0 $ is present in both cases. The set of all distinct solutions is:
$ \left\{ -\frac{\pi}{3}, -\frac{4\pi}{9}, -\frac{2\pi}{9}, 0, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{\pi}{3} \right\} $There are 7 distinct solutions in the given interval $ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $.
Similar Questions
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Consider the following two statements :-
Statement p :
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Statement q :
The angles A, B, C and D of any quadrilateral ABCD satisfy the equation $cos \left( \frac{1}{2}(A+C) \right) + cos \left( \frac{1}{2}(B+D) \right) = 0$
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Important Questions from Trigonometry
A line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at A and B such that $(PA).(PB)$ is maximum. Then $5(PA^2 + PB^2)$ is equal to :
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- The number of solutions of the equation $(4-\sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}, x \in [-2\pi, \frac{5\pi}{2}]$ is
Consider the following two statements :-
Statement p :
The value of $sin120^\circ$ can be derived by taking $\theta = 240^\circ$ in the equation $2sin \frac{\theta}{2} = \sqrt{1+sin \theta} - \sqrt{1-sin \theta}$
Statement q :
The angles A, B, C and D of any quadrilateral ABCD satisfy the equation $cos \left( \frac{1}{2}(A+C) \right) + cos \left( \frac{1}{2}(B+D) \right) = 0$
Then the truth values of p and q are respectively :-
