Solveeit Logo
Login

Question

Question: 3COS2X- 5COS CUBE X = -2 THE VLAE OF X BELONGS TIO X TO NPIE/ 2 IF THRE SRE 7 DUSTINCT SOLN WHST IUS...

3COS2X- 5COS CUBE X = -2 THE VLAE OF X BELONGS TIO X TO NPIE/ 2 IF THRE SRE 7 DUSTINCT SOLN WHST IUS GETST VSLEUE OF N

A

3

B

4

C

5

D

6

Answer

5

Explanation

Solution

The given trigonometric equation is 3cos⁡(2x)−5cos⁡3(x)=−23\cos(2x) - 5\cos^3(x) = -2. Using cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1, we get 3(2cos⁡2(x)−1)−5cos⁡3(x)=−23(2\cos^2(x) - 1) - 5\cos^3(x) = -2, which simplifies to 5cos⁡3(x)−6cos⁡2(x)+1=05\cos^3(x) - 6\cos^2(x) + 1 = 0. Let y=cos⁡(x)y = \cos(x). The equation becomes 5y3−6y2+1=05y^3 - 6y^2 + 1 = 0. The roots are y=1y=1, y=1+2110y = \frac{1+\sqrt{21}}{10}, and y=1−2110y = \frac{1-\sqrt{21}}{10}. Let α=arccos⁡(1+2110)\alpha = \arccos\left(\frac{1+\sqrt{21}}{10}\right) and β=arccos⁡(1−2110)\beta = \arccos\left(\frac{1-\sqrt{21}}{10}\right). The distinct solutions in [0,2π][0, 2\pi] are 0,α,β,2π−β,2π−α,2π0, \alpha, \beta, 2\pi-\beta, 2\pi-\alpha, 2\pi. For 7 distinct solutions in [0,nπ/2][0, n\pi/2], the interval must include 2π+α2\pi+\alpha. So, 2π+α≤nπ/2<2π+β2\pi+\alpha \le n\pi/2 < 2\pi+\beta. This implies 4+2απ≤n<4+2βπ4 + \frac{2\alpha}{\pi} \le n < 4 + \frac{2\beta}{\pi}. Given 0<α<π/30 < \alpha < \pi/3 and π/2<β<2π/3\pi/2 < \beta < 2\pi/3, we have 4<4+2απ<4.674 < 4 + \frac{2\alpha}{\pi} < 4.67 and 5<4+2βπ<5.345 < 4 + \frac{2\beta}{\pi} < 5.34. Thus, 4.something≤n<5.something4.something \le n < 5.something. The only integer value for nn is 5.