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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

The general solution of the equation 3-5 x+ ^2 x + x=0 is

The general solution of the equation 3−5sin⁡x+sin⁡2x+cos⁡x=0\sqrt{3-5 \sin x+\sin ^2 x}+\cos x=0 is

Options

  1. Anπ+(−1)nπ6,n∈Zn \pi+(-1)^n \frac{\pi}{6}, n \in Z
  2. B2nπ±π6,n∈Z2 n \pi \pm \frac{\pi}{6}, n \in Z
  3. C(2n+1)π−π6,n∈Z(2 n+1) \pi-\frac{\pi}{6}, n \in Z
  4. D2nπ±5π6,n∈Z2 n \pi \pm \frac{5 \pi}{6}, n \in Z

Correct answer

C. (2n+1)π−π6,n∈Z(2 n+1) \pi-\frac{\pi}{6}, n \in Z

Step-by-step solution

Given equation, 3−5sin⁡x+sin⁡2x+\cosx=0⇒3−5sin⁡x+sin⁡2x=−\cosx,{cos⁡x}Onsquaringbothsides,weget\begin{gathered}\sqrt{3-5 \sin x+\sin ^2 x}+\cosx=0 \\ \Rightarrow \quad \sqrt{3-5 \sin x+\sin ^2 x}=-\cosx,\{\cos x\} \end{gathered} On squaring both sides, we get⇒3−5sin⁡x+sin⁡2x=cos⁡2x⇒2sin⁡2x−5sin⁡x+2=0⇒2sin⁡2x−4sin⁡x−sin⁡x+2=0⇒2sin⁡x(sin⁡x−2)−1(sin⁡x−2)=0⇒sin⁡x=12 or 2∵sin⁡x∈[−1,1]∴sin⁡x=12 and \cosx<0\begin{array}{lrl} \Rightarrow& 3-5 \sin x+\sin ^2 x=\cos ^2 x \\ \Rightarrow & 2 \sin ^2 x-5 \sin x+2=0 \\ \Rightarrow & 2 \sin ^2 x-4 \sin x-\sin x+2=0 \\ \Rightarrow & 2 \sin x(\sin x-2)-1(\sin x-2)=0 \\ \Rightarrow & \sin x=\frac{1}{2} \text { or }2 \\ \because & \sin x \in[-1,1] \\ \therefore & \sin x=\frac{1}{2} \text { and } \cosx < 0 \end{array}Means,Means,\quad \sin x=\frac{1}{2}andand\cos x=\frac{-\sqrt{3}}{2}Therefore,Therefore,x=(2 n+1) \pi-\frac{\pi}{6}, n \in Z$.
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