According to any general solution of trigonometric equation ,

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

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.

\[\left( \mathbf{i} \right)~si{{n}^{2}}~x\text{ }\text{ }cos\text{ }x\text{ }=\text{ }1/4\]

Or,

\[si{{n}^{2}}~x\text{ }\text{ }cos\text{ }x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\]

\[1\text{ }\text{ }co{{s}^{2}}~x\text{ }\text{ }cos\text{ }x\text{ }=\text{ }1/4\]

\[\left[ as,\text{ }si{{n}^{2}}~x\text{ }=\text{ }1\text{ }\text{ }co{{s}^{2}}~x \right]\] Or,

\[4\text{ }\text{ }4\text{ }co{{s}^{2}}~x\text{ }\text{ }4\text{ }cos\text{ }x\text{ }=\text{ }1\]

Or,

\[4co{{s}^{2}}~x\text{ }+\text{ }4cos\text{ }x\text{ }\text{ }3\text{ }=\text{ }0\]

Let cos x be ‘k’

So,

\[4{{k}^{2}}~+\text{ }4k\text{ }\text{ }3\text{ }=\text{ }0\]

Or,

\[4{{k}^{2}}~\text{ }2k\text{ }+\text{ }6k\text{ }\text{ }3\text{ }=\text{ }0\]

Or,

\[2k\text{ }\left( 2k\text{ }\text{ }1 \right)\text{ }+\text{ }3\text{ }\left( 2k\text{ }\text{ }1 \right)\text{ }=\text{ }0\]

\[\left( 2k\text{ }\text{ }1 \right)\text{ }+\text{ }\left( 2k\text{ }+\text{ }3 \right)\text{ }=\text{ }0\]

Or,

\[\left( 2k\text{ }\text{ }1 \right)\text{ }=\text{ }0\text{ }or\text{ }\left( 2k\text{ }+\text{ }3 \right)\text{ }=\text{ }0\]

Or,

\[k\text{ }=\text{ }1/2\text{ }or\text{ }k\text{ }=\text{ }-3/2\]

\[cos\text{ }x\text{ }=\text{ }1/2\]

or

\[~cos\text{ }x\text{ }=\text{ }-3/2\]

we shall consider only

\[cos\text{ }x\text{ }=\text{ }1/2.\text{ }cos\text{ }x\text{ }=\text{ }-3/2\]

is not possible.

so,

\[cos\text{ }x\text{ }=\text{ }cos\text{ }{{60}^{o}}~=\text{ }cos\text{ }\pi /3\]

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

∴ the general solution is

\[x\text{ }=\text{ }2n\pi \text{ }\pm \text{ }\pi /3\] , where n ϵ Z.

\[\left( \mathbf{ii} \right)~2\text{ }co{{s}^{2}}~x\text{ }\text{ }5\text{ }cos\text{ }x\text{ }+\text{ }2\text{ }=\text{ }0\]

Or,

\[2\text{ }co{{s}^{2}}~x\text{ }\text{ }5\text{ }cos\text{ }x\text{ }+\text{ }2\text{ }=\text{ }0\]

Let cos x be ‘k’

\[2{{k}^{2}}~\text{ }5k\text{ }+\text{ }2\text{ }=\text{ }0\]

Or,

\[2{{k}^{2}}~\text{ }4k\text{ }\text{ }k\text{ }+2\text{ }=\text{ }0\]

\[2k\left( k\text{ }\text{ }2 \right)\text{ }-1\left( k\text{ }-2 \right)\text{ }=\text{ }0\]

Or,

\[\left( k\text{ }\text{ }2 \right)\text{ }\left( 2k\text{ }\text{ }1 \right)\text{ }=\text{ }0\]

\[k\text{ }=\text{ }2\text{ }or\text{ }k\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]

or,

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

we shall consider only

\[cos\text{ }x\text{ }=\text{ }1/2.\text{ }cos\text{ }x\text{ }=\text{ }2\] is not possible.

so,

\[cos\text{ }x\text{ }=\text{ }cos\text{ }{{60}^{o}}~=\text{ }cos\text{ }\pi /3\]

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

∴ the general solution is

\[x\text{ }=\text{ }2n\pi \text{ }\pm \text{ }\pi /3\] , where n ϵ Z.