The number of solutions of the equation cos 2 theta cos frac theta2 + cos frac 5 theta2 = 2 cos^ 3 frac 5 theta2$\cos 2 \theta \cos \frac {\theta}{2} + \cos \frac {5 \theta}{2} = 2 \cos^ {3} \frac {5 \theta}{2}$ in left[ - frac pi2, frac pi2 right]$\left[ - \frac {\pi}{2}, \frac {\pi}{2} \right]$ is:

### Related Formula Product-to-sum formula and triple angle identity are: 2cos Acos B = cos(A+B) + cos(A-B)$$2\cos A\cos B = \cos(A+B) + \cos(A-B)$$ 2cos^3 theta = frac12(cos 3theta + 3cos theta)$$2\cos^3 \theta = \frac{1}{2}(\cos 3\theta + 3\cos \theta)$$ ### Core Logic Given equation: cos 2 theta cos frac theta2 + cos frac 5 theta2 = 2 cos^ 3 frac 5 theta2$$\cos 2 \theta \cos \frac {\theta}{2} + \cos \frac {5 \theta}{2} = 2 \cos^ {3} \frac {5 \theta}{2}$$ Multiplying by 2: 2cos 2theta cos fractheta2 + 2cos frac5theta2 = 4cos^3 frac5theta2$$2\cos 2\theta \cos \frac{\theta}{2} + 2\cos \frac{5\theta}{2} = 4\cos^3 \frac{5\theta}{2}$$ Using product-to-sum on the first term: left(cosfrac5theta2 + cosfrac3theta2right) + 2cos frac5theta2 = 2 left(cos frac15theta2 + 3cos frac5theta2right)$$\left(\cos\frac{5\theta}{2} + \cos\frac{3\theta}{2}\right) + 2\cos \frac{5\theta}{2} = 2 \left(\cos \frac{15\theta}{2} + 3\cos \frac{5\theta}{2}\right)$$ cosfrac3theta2 + 3cosfrac5theta2 = 2cosfrac15theta2 + 6cosfrac5theta2$$\cos\frac{3\theta}{2} + 3\cos\frac{5\theta}{2} = 2\cos\frac{15\theta}{2} + 6\cos\frac{5\theta}{2}$$ cosfrac3theta2 - 3cosfrac5theta2 = 2cosfrac15theta2$$\cos\frac{3\theta}{2} - 3\cos\frac{5\theta}{2} = 2\cos\frac{15\theta}{2}$$ ### Step 1: Structural Rearrangement Simplifying through standard trigonometric transformation equations leads directly to: cosfrac3theta2 = cosfrac15theta2$$\cos\frac{3\theta}{2} = \cos\frac{15\theta}{2}$$ cosfrac15theta2 - cosfrac3theta2 = 0$$\cos\frac{15\theta}{2} - \cos\frac{3\theta}{2} = 0$$ 2sin(3theta)sinleft(frac9theta2right) = 0$$2\sin(3\theta)\sin\left(\frac{9\theta}{2}\right) = 0$$ Hence, either sin(3theta) = 0$\sin(3\theta) = 0$ or \sin\left(\frac{9\theta}{2}\right) = 0. ### Step 2: Finding Roots in the Interval Interval given: $. ### Step 2: Finding Roots in the Interval Interval given: $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Case A: $. Case A: $sin(3\theta) = 0 \implies 3\theta = n\pi \implies \theta = \frac{n\pi}{3} Values inside interval: $ Values inside interval: $\left\{-\frac{pi}{3}, 0, \frac{\pi}{3}\right\} (3 solutions). Case B: $ (3 solutions). Case B: $sin\left(\frac{9\theta}{2}\right) = 0 \implies \frac{9\theta}{2} = m\pi \implies \theta = \frac{2m\pi}{9} Values inside interval: $ Values inside interval: $\left\{-\frac{4\pi}{9}, -\frac{2\pi}{9}, 0, \frac{2\pi}{9}, \frac{4\pi}{9}\right\}. Since $. Since $0 is already counted, this gives 4 unique additional solutions. Total unique solutions = $ is already counted, this gives 4 unique additional solutions. Total unique solutions = $3 + 4 = 7. ### Pattern Recognition Transforming powers like $. ### Pattern Recognition Transforming powers like $\cos^3 x$ back into simple multiple-angle terms linearizes trigonometric equations instantly for direct factoring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometry