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Home » Find the general solutions of the following equations: (iii) sin 9x = sin x (iv) sin 2x = cos 3x
Find the general solutions of the following equations: (iii) sin 9x = sin x (iv) sin 2x = cos 3x
CBSE | Class 11 | Exercise 11.1 | Maths | RD Sharma | Trigonometric Equations
Find the general solutions of the following equations: (iii) sin 9x = sin x (iv) sin 2x = cos 3x

    \[~\left( \mathbf{iii} \right)~sin\text{ }9x\text{ }=\text{ }sin\text{ }x\]

Or,

    \[Sin\text{ }9x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }0\]

Using transformation 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,

    \[=\text{ }2\text{ }cos\text{ }\left( 9x+x \right)/2\text{ }sin\text{ }\left( 9x-x \right)/2\]

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

    \[Cos\text{ }5x\text{ }=\text{ }0\text{ }or\text{ }sin\text{ }4x\text{ }=\text{ }0\]

Let us verify both the expressions,

    \[Cos\text{ }5x\text{ }=\text{ }0\]

    \[Cos\text{ }5x\text{ }=\text{ }cos\text{ }\pi /2\]

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

    \[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\pi /10,\]

where n ϵ Z.

    \[sin\text{ }4x\text{ }=\text{ }0\]

    \[sin\text{ }4x\text{ }=\text{ }sin\text{ }0\]

    \[sin\text{ }4x\text{ }=\text{ }sin\text{ }0\]

    \[4x\text{ }=\text{ }n\pi \]

    \[x\text{ }=\text{ }n\pi /4,\]

where n ϵ Z.

∴ the general solution is

    \[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\pi /10\text{ }or\text{ }n\pi /4,\]

where n ϵ Z.

    \[\left( \mathbf{iv} \right)~sin\text{ }2x\text{ }=\text{ }cos\text{ }3x\]

Or,

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

or,

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

    \[\left[ as,\text{ }sin\text{ }A\text{ }=\text{ }cos\text{ }\left( \pi /2\text{ }\text{ }A \right) \right]\]

    \[\pi /2\text{ }\text{ }2x\text{ }=\text{ }2n\pi \text{ }\pm \text{ }3x\]

or,

    \[\pi /2\text{ }\text{ }2x\text{ }=\text{ }2n\pi \text{ }+\text{ }3x\]

[or]

    \[\pi /2\text{ }\text{ }2x\text{ }=\text{ }2n\pi \text{ }\text{ }3x\pi /2\text{ }\text{ }2x\text{ }=\text{ }2n\pi \text{ }\text{ }3x\]

    \[5x\text{ }=\text{ }\pi /2\text{ }+\text{ }2n\pi \]

[or]

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

Or,

    \[5x\text{ }=\text{ }\pi /2\text{ }\left( 1\text{ }+\text{ }4n \right)\]

[or]

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

    \[x\text{ }=\text{ }\pi /10\text{ }\left( 1\text{ }+\text{ }4n \right)\]

[or]

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

or,

∴ the general solution is

    \[x\text{ }=\text{ }\pi /10\text{ }\left( 4n\text{ }+\text{ }1 \right)\]

or,

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

where n ϵ Z.

i

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