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Home » Find the general solutions of the following equations: (i) sin 2x = √3/2 (ii) cos 3x = 1/2
Find the general solutions of the following equations: (i) sin 2x = √3/2 (ii) cos 3x = 1/2
CBSE | Class 11 | Exercise 11.1 | Maths | RD Sharma | Trigonometric Equations
Find the general solutions of the following equations: (i) sin 2x = √3/2 (ii) cos 3x = 1/2

The general solution of any trigonometric equation is given as:

    \[sin\text{ }x\text{ }=\text{ }sin\text{ }y,\]

or,

    \[x\text{ }=\text{ }n\pi \text{ }+\text{ }{{\left( \text{ }1 \right)}^{n~}}y,\]

where n ∈ Z.

    \[cos\text{ }x\text{ }=\text{ }cos\text{ }y,\]

or,

    \[x\text{ }=\text{ }2n\pi ~\pm ~y,\]

where n ∈ Z.

    \[tan\text{ }x\text{ }=\text{ }tan\text{ }y,\]

or,

    \[x\text{ }=\text{ }n\pi ~+\text{ }y,\]

where n ∈ Z.

    \[\left( \mathbf{i} \right)~sin\text{ }2x\text{ }=\text{ }\surd 3/2\]

or,

    \[sin\text{ }2x\text{ }=\text{ }\surd 3/2\]

    \[=\text{ }sin\text{ }\left( \pi /3 \right)\]

∴ the general solution is

    \[2x\text{ }=\text{ }n\pi \text{ }+\text{ }{{\left( -1 \right)}^{n}}~\pi /3,\]

where n ϵ Z.

    \[x\text{ }=\text{ }n\pi /2\text{ }+\text{ }{{\left( -1 \right)}^{n}}~\pi /6,\]

where n ϵ Z.

    \[\left( \mathbf{ii} \right)~cos\text{ }3x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]

or,

    \[cos\text{ }3x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]

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

∴ the general solution is

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

where n ϵ Z.

    \[x\text{ }=\text{ }2n\pi /3\text{ }\pm \text{ }\pi /9,\]

where n ϵ Z.

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