Find the general solutions of the following equations: (ix) tan px = cot qx (x) sin 2x + cos x = 0
Find the general solutions of the following equations: (ix) tan px = cot qx (x) sin 2x + cos x = 0

\[\left( \mathbf{ix} \right)~tan\text{ }px\text{ }=\text{ }cot\text{ }qx\]

Or,

\[tan\text{ }px\text{ }=\text{ }cot\text{ }qx\]

or,

\[tan\text{ }px\text{ }=\text{ }tan\text{ }\left( \pi /2\text{ }\text{ }qx \right)\]

\[~\left[ as,\text{ }cot\text{ }A\text{ }=\text{ }tan\text{ }\left( \pi /2\text{ }\text{ }A \right) \right]\]

\[px\text{ }=\text{ }n\pi \text{ }\pm \text{ }\left( \pi /2\text{ }\text{ }qx \right)\]

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

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

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

∴ the general solution is

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

where n ϵ Z.

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

Or,

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

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

Or,

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

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

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

\[~\left[ as,\text{ }-cos\text{ }A\text{ }=\text{ }cos\text{ }\left( \pi \text{ }\text{ }A \right) \right]\]

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

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

So,

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

[or]

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

or,

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

[or]

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

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

[or]

\[x\text{ }=\text{ }\pi /6\text{ }\left( 4n\text{ }\text{ }1 \right)\]

∴ the general solution is

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

where n ϵ Z.

\[~x\text{ }=\text{ }\pi /6\text{ }\left( 4n\text{ }\text{ }1 \right)\]

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

where n ϵ Z.

\[~x\text{ }=\text{ }\pi /6\text{ }\left( 4n\text{ }\text{ }1 \right),\]

where n ϵ Z.