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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Trigonometric Equations

BITSAT 2024 · Mathematics · Trigonometric Equations

Number of solutions of equations 9 θ= θ in the interval [0,2 π] is

Number of solutions of equations sin⁡9θ=sin⁡θ\sin 9 \theta=\sin \theta in the interval [0,2π][0,2 \pi] is

Options

  1. A16
  2. B17
  3. C18
  4. D15

Correct answer

B. 17

Step-by-step solution

 Given, sin⁡9θ=sin⁡θ⇒sin⁡9θ−sin⁡θ=0⇒2cos⁡(9θ+θ2)sin⁡(9θ−θ2)=0[∵sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)]⇒2cos⁡5θsin⁡4θ=0⇒ Either cos⁡5θ=0 or  sin⁡4θ=0, also θ∈[0,2π]⇒5θ=(2n+1)π2 or 4θ=nπ⇒θ=(2n+1)π10 or θ=nπ4∴θ=π10,3π10,π2,7π10,9π10,11π10,13π10,3π2,17π10,19π10 or 0,π4,π2,3π4,π,5π4,3π2,7π4,2π∴ Total number of solutions =17(∵π2 and 3π2 are common )\begin{aligned} & \text{ Given, } \quad \sin 9 \theta=\sin \theta \\ & \Rightarrow \quad \sin 9 \theta-\sin \theta=0 \\ & \Rightarrow \quad 2 \cos \left(\frac{9 \theta+\theta}{2}\right) \sin \left(\frac{9 \theta-\theta}{2}\right)=0 \\ & {\left[\because \sin C-\sin D=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]} \\ & \Rightarrow \quad 2 \cos 5 \theta \sin 4 \theta=0 \\ & \Rightarrow \text{ Either } \cos 5 \theta=0 \text{\text{ or } } \sin 4 \theta=0 \text{, also } \theta \in[0,2 \pi] \\ & \Rightarrow \quad 5 \theta=(2 n+1) \frac{\pi}{2} \\ & \text{ or } \quad 4 \theta=n \pi \\ & \Rightarrow \quad \theta=(2 n+1) \frac{\pi}{10} \text{ or } \theta=\frac{n \pi}{4} \\ & \therefore \quad \theta=\frac{\pi}{10}, \frac{3 \pi}{10}, \frac{\pi}{2}, \frac{7 \pi}{10}, \frac{9 \pi}{10}, \frac{11 \pi}{10}, \frac{13 \pi}{10}, \frac{3 \pi}{2}, \frac{17 \pi}{10}, \frac{19 \pi}{10} \\ & \text{ or } \quad 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi, \frac{5 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}, 2 \pi \\ & \therefore \text{ Total number of solutions }=17 \\ & \left(\because \frac{\pi}{2} \text{ and } \frac{3 \pi}{2} \text{ are common }\right) \end{aligned}
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