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Home » Solve the following equations: (i) tan x + tan 2x + tan 3x = 0 (ii) tan x + tan 2x = tan 3x
Solve the following equations: (i) tan x + tan 2x + tan 3x = 0 (ii) tan x + tan 2x = tan 3x
CBSE | Class 11 | Exercise 11.1 | Maths | RD Sharma | Trigonometric Equations
Solve the following equations: (i) tan x + tan 2x + tan 3x = 0 (ii) tan x + tan 2x = tan 3x

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.

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

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

where n ∈ Z.

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

or,

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

where n ∈ Z.

    \[\left( \mathbf{i} \right)~tan\text{ }x\text{ }+\text{ }tan\text{ }2x\text{ }+\text{ }tan\text{ }3x\text{ }=\text{ }0\]

Or,

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

    \[tan\text{ }x\text{ }+\text{ }tan\text{ }2x\text{ }+\text{ }tan\text{ }\left( x\text{ }+\text{ }2x \right)\text{ }=\text{ }0\]

By using the formula,

    \[tan\text{ }\left( A+B \right)\text{ }=\text{ }\left[ tan\text{ }A\text{ }+\text{ }tan\text{ }B \right]\text{ }/\text{ }\left[ 1\text{ }\text{ }tan\text{ }A\text{ }tan\text{ }B \right]\]

So,

    \[tan\text{ }x\text{ }+\text{ }tan\text{ }2x\text{ }+\text{ }\left[ \left[ tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right]/ \right[1-\text{ }tan\text{ }x\text{ }tan\text{ }2x]]\text{ }=\text{ }0\]

    \[tan\text{ }x\text{ }+\text{ }tan\text{ }2x\text{ }+\text{ }\left[ \left[ tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right]/ \right[1-\text{ }tan\text{ }x\text{ }tan\text{ }2x]]\text{ }=\text{ }0\]

    \[\left( tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right)\text{ }\left( 1\text{ }+\text{ }1/\left( 1-\text{ }tan\text{ }x\text{ }tan\text{ }2x \right) \right)\text{ }=\text{ }0\]

Or,

    \[\left( tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right)\text{ }\left( \left[ 2\text{ }\text{ }tan\text{ }x\text{ }tan\text{ }2x \right]\text{ }/\text{ }\left[ 1\text{ }\text{ }tan\text{ }x\text{ }tan\text{ }2x \right] \right)\text{ }=\text{ }0\]

Then,

    \[\left( tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right)\text{ }=\text{ }0\text{ }or\text{ }\left( \left[ 2\text{ }\text{ }tan\text{ }x\text{ }tan\text{ }2x \right]\text{ }/\text{ }\left[ 1\text{ }\text{ }tan\text{ }x\text{ }tan\text{ }2x \right] \right)\text{ }=\text{ }0\]

    \[\left( tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right)\text{ }=\text{ }0\text{ }or\text{ }\left[ 2\text{ }\text{ }tan\text{ }x\text{ }tan\text{ }2x \right]\text{ }=\text{ }0\]

Or,

    \[tan\text{ }x\text{ }=\text{ }tan\text{ }\left( -2x \right)\text{ }or\text{ }tan\text{ }x\text{ }tan\text{ }2x\text{ }=\text{ }2\]

    \[x\text{ }=\text{ }n\pi \text{ }+\text{ }\left( -2x \right)\text{ }or\text{ }tax\text{ }x\text{ }\left[ 2tan\text{ }x/\left( 1\text{ }\text{ }ta{{n}^{2}}~x \right) \right]\text{ }=\text{ }2\]

    \[~\left[ Using,\text{ }tan\text{ }2x\text{ }=\text{ }2\text{ }tan\text{ }x\text{ }/\text{ }1-ta{{n}^{2}}~x \right]\]

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

Or,

    \[3x\text{ }=\text{ }n\pi \text{ }or\text{ }2\text{ }ta{{n}^{2}}~x\text{ }=\text{ }2\left( 1\text{ }\text{ }ta{{n}^{2}}~x \right)\]

    \[3x\text{ }=\text{ }n\pi \text{ }or\text{ }2\text{ }ta{{n}^{2}}~x\text{ }=\text{ }2\text{ }\text{ }2ta{{n}^{2}}~x\]

Or,

    \[3x\text{ }=\text{ }n\pi \]

or

    \[4\text{ }ta{{n}^{2}}~x\text{ }=\text{ }2\]

    \[x\text{ }=\text{ }n\pi /3\]

or

    \[ta{{n}^{2}}~x\text{ }=\text{ }2/4\]

    \[x\text{ }=\text{ }n\pi /3\]

or

    \[~ta{{n}^{2}}~x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]

    \[x\text{ }=\text{ }n\pi /3\]

or

    \[tan\text{ }x\text{ }=\text{ }1/\surd 2\]

    \[x\text{ }=\text{ }n\pi /3\]

or

    \[~x\text{ }=\text{ }tan\text{ }\alpha \]

    \[\left[ let\text{ }1/\surd 2\text{ }be\text{ }\alpha \right]\]

    \[x\text{ }=\text{ }n\pi /3\]

or

    \[x\text{ }=\text{ }m\pi \text{ }+\text{ }\alpha \]

∴ the general solution is

    \[x\text{ }=\text{ }n\pi /3\text{ }or\text{ }m\pi \text{ }+\text{ }\alpha ,\]

where

    \[\alpha \text{ }=\text{ }ta{{n}^{-1}}1/\surd 2,\text{ }m,\text{ }n\in ~Z.\]

    \[\left( \mathbf{ii} \right)~tan\text{ }x\text{ }+\text{ }tan\text{ }2x\text{ }=\text{ }tan\text{ }3x\]

Or,

    \[tan\text{ }x\text{ }+\text{ }tan\text{ }2x\text{ }=\text{ }tan\text{ }3x\]

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

or,

    \[tan\text{ }x\text{ }+\text{ }tan\text{ }2x\text{ }\text{ }tan\text{ }\left( x\text{ }+\text{ }2x \right)\text{ }=\text{ }0\]

By using the formula,

    \[tan\text{ }\left( A+B \right)\text{ }=\text{ }\left[ tan\text{ }A\text{ }+\text{ }tan\text{ }B \right]\text{ }/\text{ }\left[ 1\text{ }\text{ }tan\text{ }A\text{ }tan\text{ }B \right]\]

So,

    \[tan\text{ }x\text{ }+\text{ }tan\text{ }2x\text{ }\text{ }\left[ \left[ tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right]/ \right[1-\text{ }tan\text{ }x\text{ }tan\text{ }2x]]\text{ }=\text{ }0\]

    \[\left( tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right)\text{ }\left( 1\text{ }\text{ }1/\left( 1-\text{ }tan\text{ }x\text{ }tan\text{ }2x \right) \right)\text{ }=\text{ }0\]

Or,

    \[\left( tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right)\text{ }\left( \left[ \text{ }tan\text{ }x\text{ }tan\text{ }2x \right]\text{ }/\text{ }\left[ 1\text{ }\text{ }tan\text{ }x\text{ }tan\text{ }2x \right] \right)\text{ }=\text{ }0\]

Then,

    \[\left( tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right)\text{ }=\text{ }0\]

or

    \[\left( \left[ \text{ }tan\text{ }x\text{ }tan\text{ }2x \right]\text{ }/\text{ }\left[ 1\text{ }\text{ }tan\text{ }x\text{ }tan\text{ }2x \right] \right)\text{ }=\text{ }0\]

Or,

    \[\left( tan\text{ }x\text{ }+\text{ }tan\text{ }2x \right)\text{ }=\text{ }0\text{ }or\text{ }\left[ \text{ }tan\text{ }x\text{ }tan\text{ }2x \right]\text{ }=\text{ }0\]

Or,

    \[tan\text{ }x\text{ }=\text{ }tan\text{ }\left( -2x \right)\text{ }or\text{ }-tan\text{ }x\text{ }tan\text{ }2x\text{ }=\text{ }0\]

or,

    \[tan\text{ }x\text{ }=\text{ }tan\text{ }\left( -2x \right)\text{ }or\text{ }2ta{{n}^{2}}~x\text{ }/\text{ }\left( 1\text{ }\text{ }ta{{n}^{2}}~x \right)\text{ }=\text{ }0\]

    \[~\left[ Using,\text{ }tan\text{ }2x\text{ }=\text{ }2\text{ }tan\text{ }x\text{ }/\text{ }1-ta{{n}^{2}}~x \right]\]

    \[x\text{ }=\text{ }n\pi \text{ }+\text{ }\left( -2x \right)\text{ }or\text{ }x\text{ }=\text{ }m\pi \text{ }+\text{ }0\]

or,

    \[3x\text{ }=\text{ }n\pi \text{ }or\text{ }x\text{ }=\text{ }m\pi \]

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

∴ the general solution is

    \[x\text{ }=\text{ }n\pi /3\text{ }or\text{ }m\pi ,\text{ }where\text{ }m,\text{ }n\in ~Z.\]

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