NIMCET 2018 Question Paper with solutions

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Question : 2 of 120

Marks: +1, -0

The number of the solutions of the equation

\sin x+\sin 5 x=\sin 3 x

, lying in the interval

[0, π]

, is:

4
6
5
2

Solution:

\sin x+\sin 5 x=\sin 3 x

Using

\sin 2 A+\sin 2 B=2 \sin (A+B) \cos (A-B)

, we get:

2 \sin 3 x \cos 2 x=\sin 3 x

⇒ \sin 3 x(2 \cos 2 x-1)=0

⇒ \sin 3 x=0

OR

2 \cos 2 x-1=0

CASE

1: \sin 3 x=0=\sin n π, n ∈ Z .

⇒ x=\;{n π}/{3}

⇒ x=. . ., 0, \;{π}/{3}, \;{2 π}/{3}, π, . . .

CASE

2: 2 \cos 2 x-1=0

⇒ \cos 2 x=\;{1}/{2}=\cos (2 n π ± \;{π}/{3}), n ∈ Z .

⇒ x=n π ± \;{π}/{6}=(6 n ± 1) \;{π}/{6}

⇒ x=. . ., \;{π}/{6}, \;{5 π}/{6}, . . .

Therefore, there are 6 possible values of

x

in the interval

[0, π]

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