The number of solutions of the equation cos⁡2θcos⁡θ2+cos⁡5θ2=2cos⁡35θ2\cos 2\theta\cos\frac{\theta}{2} + \cos\frac{5\theta}{2} = 2\cos^3\frac{5\theta}{2} in [−π2,π2][-\tfrac{\pi}{2},\tfrac{\pi}{2}] is:

Answer: 7

Step-by-step solution

Step 1: Rearrange the equation

The given equation is cos⁡2θcos⁡θ2+cos⁡5θ2=2cos⁡35θ2\cos 2\theta\cos\frac{\theta}{2} + \cos\frac{5\theta}{2} = 2\cos^3\frac{5\theta}{2}. We can rearrange this equation by moving the term cos⁡5θ2\cos\frac{5\theta}{2} to the right side.

Step 2: Factor the right side using the triple angle identity

We can factor out cos⁡5θ2\cos\frac{5\theta}{2} from the right side. The expression 2cos⁡2x−12\cos^2 x - 1 is the double angle identity for cos⁡2x\cos 2x. Here, x=5θ2x = \frac{5\theta}{2}, so 2cos⁡25θ2−1=cos⁡(2⋅5θ2)=cos⁡5θ2\cos^2\frac{5\theta}{2} - 1 = \cos(2 \cdot \frac{5\theta}{2}) = \cos 5\theta.

Step 3: Apply product-to-sum identities

We use the product-to-sum identity 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A+B) + \cos(A-B) on both sides of the equation. This transforms the products of cosines into sums of cosines.

Step 4: Simplify the arguments and solve

After applying the product-to-sum identities and simplifying the arguments, we get cos⁡5θ2+cos⁡3θ2=cos⁡15θ2+cos⁡5θ2\cos\frac{5\theta}{2} + \cos\frac{3\theta}{2} = \cos\frac{15\theta}{2} + \cos\frac{5\theta}{2}. The term cos⁡5θ2\cos\frac{5\theta}{2} cancels out from both sides, leaving cos⁡3θ2=cos⁡15θ2\cos\frac{3\theta}{2} = \cos\frac{15\theta}{2}. The general solution for cos⁡A=cos⁡B\cos A = \cos B is A=2nπ±BA = 2n\pi \pm B, where nn is an integer.

Step 5: Solve for θ\theta in two cases

We consider two cases based on the general solution. In Case 1, we take the positive sign, leading to θ=nπ3\theta = \frac{n\pi}{3}. In Case 2, we take the negative sign, leading to θ=2nπ9\theta = \frac{2n\pi}{9}.

Step 6: Find solutions in the interval [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

We need to find the values of θ\theta that lie in the interval [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], which is approximately [−1.57,1.57][-1.57, 1.57]. For θ=nπ3\theta = \frac{n\pi}{3}: n=0,±1n=0, \pm 1 give 0,±π30, \pm \frac{\pi}{3}. For θ=2nπ9\theta = \frac{2n\pi}{9}: n=0,±1,±2n=0, \pm 1, \pm 2 give 0,±2π9,±4π90, \pm \frac{2\pi}{9}, \pm \frac{4\pi}{9}. The unique solutions are 0,±π3,±2π9,±4π90, \pm \frac{\pi}{3}, \pm \frac{2\pi}{9}, \pm \frac{4\pi}{9}.

Step 7: Count the number of unique solutions

Listing all the unique solutions found: 0,π3,−π3,2π9,−2π9,4π9,−4π90, \frac{\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{9}, -\frac{2\pi}{9}, \frac{4\pi}{9}, -\frac{4\pi}{9}. There are 7 unique solutions.

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