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Home » Solve the following equations: (vii) sin x + sin 2x + sin 3x + sin 4x = 0 (viii) sin 3x – sin x = 4 cos2 x – 2
Solve the following equations: (vii) sin x + sin 2x + sin 3x + sin 4x = 0 (viii) sin 3x – sin x = 4 cos2 x – 2
CBSE | Class 11 | Exercise 11.1 | Maths | RD Sharma | Trigonometric Equations
Solve the following equations: (vii) sin x + sin 2x + sin 3x + sin 4x = 0 (viii) sin 3x – sin x = 4 cos2 x – 2

    \[~\left( \mathbf{vii} \right)~sin\text{ }x\text{ }+\text{ }sin\text{ }2x\text{ }+\text{ }sin\text{ }3x\text{ }+\text{ }sin\text{ }4x\text{ }=\text{ }0\]

Or,

    \[sin\text{ }x\text{ }+\text{ }sin\text{ }2x\text{ }+\text{ }sin\text{ }3x\text{ }+\text{ }sin\text{ }4x\text{ }=\text{ }0\]

using transformation formula,

    \[sin\text{ }x\text{ }+\text{ }sin\text{ }3x\text{ }+\text{ }sin\text{ }2x\text{ }+\text{ }sin\text{ }4x\text{ }=\text{ }0\]

By using the formula,

    \[sin\text{ }A\text{ }+\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }sin\text{ }\left( A+B \right)/2\text{ }cos\text{ }\left( A-B \right)/2\]

So,

    \[2\text{ }sin\text{ }\left( 3x+x \right)/2\text{ }cos\text{ }\left( 3x-x \right)/2\text{ }+\text{ }2\text{ }sin\text{ }\left( 4x+2x \right)/2\text{ }cos\text{ }\left( 4x-2x \right)/2\text{ }=\text{ }0\]

    \[2\text{ }sin\text{ }2x\text{ }cos\text{ }x\text{ }+\text{ }2\text{ }sin\text{ }3x\text{ }cos\text{ }x\text{ }=\text{ }0\]

Or,

    \[2cos\text{ }x\text{ }\left( sin\text{ }2x\text{ }+\text{ }sin\text{ }3x \right)\text{ }=\text{ }0\]

Again by using the formula,

    \[sin\text{ }A\text{ }+\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }sin\text{ }\left( A+B \right)/2\text{ }cos\text{ }\left( A-B \right)/2\]

we get,

    \[2cos\text{ }x\text{ }\left( 2\text{ }sin\text{ }\left( 3x+2x \right)/2\text{ }cos\text{ }\left( 3x-2x \right)/2 \right)\text{ }=\text{ }0\]

    \[2cos\text{ }x\text{ }\left( 2\text{ }sin\text{ }5x/2\text{ }cos\text{ }x/2 \right)\text{ }=\text{ }0\]

Or,

    \[4\text{ }cos\text{ }x\text{ }sin\text{ }5x/2\text{ }cos\text{ }x/2\text{ }=\text{ }0\]

So,

    \[Cos\text{ }x\text{ }=\text{ }0\text{ }or\text{ }sin\text{ }5x/2\text{ }=\text{ }0\text{ }or\text{ }cos\text{ }x/2\text{ }=\text{ }0\]

    \[Cos\text{ }x\text{ }=\text{ }cos\text{ }0\]

or

    \[sin\text{ }5x/2\text{ }=\text{ }sin\text{ }0\]

or

    \[cos\text{ }x/2\text{ }=\text{ }cos\text{ }0\]

    \[Cos\text{ }x\text{ }=\text{ }cos\text{ }\pi /2\]

or

    \[sin\text{ }5x/2\text{ }=\text{ }k\pi \]

or

    \[cos\text{ }x/2\text{ }=\text{ }cos\text{ }\left( 2p\text{ }+\text{ }1 \right)\text{ }\pi /2\]

    \[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /2\]

or

    \[5x/2\text{ }=\text{ }k\pi \text{ }or\text{ }x/2\text{ }=\text{ }\left( 2p\text{ }+\text{ }1 \right)\text{ }\pi /2\]

    \[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /2\]

or

    \[x\text{ }=\text{ }2k\pi /5\text{ }or\text{ }x\text{ }=\text{ }\left( 2p\text{ }+\text{ }1 \right)\]

    \[x\text{ }=\text{ }n\pi \text{ }+\text{ }\pi /2\]

or

    \[x\text{ }=\text{ }2k\pi /5\text{ }or\text{ }x\text{ }=\text{ }\left( 2p\text{ }+\text{ }1 \right)\]

∴ general solution:

    \[x\text{ }=\text{ }n\pi \text{ }+\text{ }\pi /2\]

or

    \[x\text{ }=\text{ }2k\pi /5\]

or

    \[x\text{ }=\text{ }\left( 2p\text{ }+\text{ }1 \right),\]

where n, k, p ϵ Z.

    \[\left( \mathbf{viii} \right)~sin\text{ }3x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }4\text{ }co{{s}^{2}}~x\text{ }\text{ }2\]

Or,

    \[sin\text{ }3x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }4\text{ }co{{s}^{2}}~x\text{ }\text{ }2\]

    \[sin\text{ }3x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }2\left( 2\text{ }co{{s}^{2}}~x\text{ }\text{ }1 \right)\]

or,

    \[sin\text{ }3x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }2\text{ }cos\text{ }2x\]

    \[\left[ as,~cos\text{ }2A\text{ }=\text{ }2co{{s}^{2}}~A\text{ }\text{ }1 \right]\]

By using the formula,

    \[Sin\text{ }A\text{ }\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }cos\text{ }\left( A+B \right)/2\text{ }sin\text{ }\left( A-B \right)/2\]

So,

    \[2\text{ }cos\text{ }\left( 3x+x \right)/2\text{ }sin\text{ }\left( 3x-x \right)/2\text{ }=\text{ }2\text{ }cos\text{ }2x\]

    \[2\text{ }cos\text{ }2x\text{ }sin\text{ }x\text{ }\text{ }2\text{ }cos\text{ }2x\text{ }=\text{ }0\]

Or,

    \[2\text{ }cos\text{ }2x\text{ }\left( sin\text{ }x\text{ }\text{ }1 \right)\text{ }=\text{ }0\]

Then,

    \[2\text{ }cos\text{ }2x\text{ }=\text{ }0\text{ }or\text{ }sin\text{ }x\text{ }\text{ }1\text{ }=\text{ }0\]

    \[Cos\text{ }2x\text{ }=\text{ }0\text{ }or\text{ }sin\text{ }x\text{ }=\text{ }1\]

Or,

    \[Cos\text{ }2x\text{ }=\text{ }cos\text{ }0\text{ }or\text{ }sin\text{ }x\text{ }=\text{ }sin\text{ }1\]

    \[Cos\text{ }2x\text{ }=\text{ }cos\text{ }0\text{ }or\text{ }sin\text{ }x\text{ }=\text{ }sin\text{ }\pi /2\]

Or,

    \[2x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /2\]

or

    \[x\text{ }=\text{ }m\pi \text{ }+\text{ }{{\left( -1 \right)}^{~m}}~\pi /2\]

    \[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /4\]

or

    \[x\text{ }=\text{ }m\pi \text{ }+\text{ }{{\left( -1 \right)}^{~m}}~\pi /2\]

∴ general solution:

    \[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /4\text{ }or\text{ }m\pi \text{ }+\text{ }{{\left( -1 \right)}^{~m}}~\pi /2,\]

where m, n ϵ Z.

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