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Question
The number of solutions of cos⁡2x=cos⁡x\cos 2x = \cos x in the interval [0,2π)[0, 2\pi) is _____.
Solution Path
General solution cos⁡A=cos⁡B⇒A=2nπ±B\cos A = \cos B \Rightarrow A = 2n\pi \pm B, enumerate in [0,2π)[0, 2\pi), get 3 distinct solutions.
01Question Setup
1/4
Find the number of solutions of cos⁡2x=cos⁡x\cos 2x = \cos x in [0,2π)[0, 2\pi).
Number of solutions =  ?= \;?
02General Solution
2/4
Use cos⁡A=cos⁡B⇒A=2nπ±B\cos A = \cos B \Rightarrow A = 2n\pi \pm B. So 2x=2nπ±x2x = 2n\pi \pm x. Case 1: x=2nπx = 2n\pi. Case 2: 3x=2nπ3x = 2n\pi, i.e., x=2nπ3x = \dfrac{2n\pi}{3}.
x=2nπx = 2n\pi or x=2nπ3x = \dfrac{2n\pi}{3}
03Count in [0, 2pi)KEY INSIGHT
3/4
Case 1 gives x=0x = 0. Case 2 gives x=0,2π3,4π3x = 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}. Distinct solutions: {0,2π3,4π3}\{0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}\}. Three points on the unit circle.
3 distinct solutions
04Final Answer
4/4
The number of solutions is 3.
3\boxed{3}
Concepts from this question2 concepts unlocked
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General Solution of Trig Equations

STANDARD

Standard patterns that give all solutions of basic trig equations using integer parameter n

sin⁡x=sin⁡α⇒x=nπ+(−1)nα\sin x = \sin \alpha \Rightarrow x = n\pi + (-1)^n \alpha

JEE frequently asks for the number of solutions in a given interval. Without the general solution formula, you cannot systematically list all roots.

Trig equation solvingNumber of solutions in intervalPrincipal value problems
Practice (12 Qs) →

Compound Angle Expansion

EASY

Expand sin(A+B) and cos(A+B) into products of individual trig ratios of A and B

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B

The foundation of nearly all trig manipulation in JEE. Double angle, half angle, and product formulas all derive from this. Appears in 3-4 questions every paper.

Double/half angle derivationsTrig simplificationProving identities
Practice (15 Qs) →
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