\[\left( \mathbf{iii} \right)~2\text{ }si{{n}^{2}}~x\text{ }+\text{ }\surd 3\text{ }cos\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]
Or,
\[2\text{ }si{{n}^{2}}~x\text{ }+\text{ }\surd 3\text{ }cos\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]
\[2\text{ }\left( 1\text{ }\text{ }co{{s}^{2}}~x \right)\text{ }+\text{ }\surd 3\text{ }cos\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]
Or,
\[~\left[ as,\text{ }si{{n}^{2}}~x\text{ }=\text{ }1\text{ }\text{ }co{{s}^{2}}~x \right]\]
\[2\text{ }\text{ }2\text{ }co{{s}^{2}}~x\text{ }+\text{ }\surd 3\text{ }cos\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]
or,
\[2\text{ }co{{s}^{2}}~x\text{ }\text{ }\surd 3\text{ }cos\text{ }x\text{ }\text{ }3\text{ }=\text{ }0\]
Let cos x be ‘k’
\[2{{k}^{2}}~\text{ }\surd 3\text{ }k\text{ }\text{ }3\text{ }=\text{ }0\]
Or,
\[2{{k}^{2}}~-2\surd 3\text{ }k\text{ }+\text{ }\surd 3\text{ }k\text{ }\text{ }3\text{ }=\text{ }0\]
\[2k\left( k\text{ }\text{ }\surd 3 \right)\text{ }+\surd 3\left( k\text{ }\text{ }\surd 3 \right)\text{ }=\text{ }0\]
Or,
\[\left( 2k\text{ }+\text{ }\surd 3 \right)\text{ }\left( k\text{ }\text{ }\surd 3 \right)\text{ }=\text{ }0\]
\[k\text{ }=\text{ }\surd 3\text{ }or\text{ }k\text{ }=\text{ }-\surd 3/2\]
or,
\[cos\text{ }x\text{ }=\text{ }\surd 3\text{ }or\text{ }cos\text{ }x\text{ }=\text{ }-\surd 3/2\]
we will only consider
\[cos\text{ }x\text{ }=\text{ }-\surd 3/2.\text{ }cos\text{ }x\text{ }=\text{ }\surd 3\]
is not possible.
so,
\[cos\text{ }x\text{ }=\text{ }-\surd 3/2\]
\[cos\text{ }x\text{ }=\text{ }cos\text{ }150{}^\circ \text{ }=\text{ }cos\text{ }5\pi /6\]
\[x\text{ }=\text{ }2n\pi \text{ }\pm \text{ }5\pi /6\] , where n ϵ Z.
\[\left( \mathbf{iv} \right)~4\text{ }si{{n}^{2}}~x\text{ }\text{ }8\text{ }cos\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]
Or,
\[4\text{ }si{{n}^{2}}~x\text{ }\text{ }8\text{ }cos\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]
\[4\text{ }\left( 1\text{ }\text{ }co{{s}^{2}}~x \right)\text{ }\text{ }8\text{ }cos\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]
\[\left[ as,\text{ }si{{n}^{2}}~x\text{ }=\text{ }1\text{ }\text{ }co{{s}^{2}}~x \right]\]
\[4\text{ }\text{ }4\text{ }co{{s}^{2}}~x\text{ }\text{ }8\text{ }cos\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]
Or,
\[4\text{ }co{{s}^{2}}~x\text{ }+\text{ }8\text{ }cos\text{ }x\text{ }\text{ }5\text{ }=\text{ }0\]
Let cos x be ‘k’
\[4{{k}^{2}}~+\text{ }8k\text{ }\text{ }5\text{ }=\text{ }0\]
\[4{{k}^{2}}~-2k\text{ }+\text{ }10k\text{ }\text{ }5\text{ }=\text{ }0\]
Or,
\[2k\left( 2k\text{ }\text{ }1 \right)\text{ }+\text{ }5\left( 2k\text{ }\text{ }1 \right)\text{ }=\text{ }0\]
\[\left( 2k\text{ }+\text{ }5 \right)\text{ }\left( 2k\text{ }\text{ }1 \right)\text{ }=\text{ }0\]
\[k\text{ }=\text{ }-5/2\text{ }=\text{ }-2.5\text{ }or\text{ }k\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]
\[cos\text{ }x\text{ }=\text{ }-2.5\text{ }or\text{ }cos\text{ }x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]
we shall consider only
\[cos\text{ }x\text{ }=\text{ }1/2.\text{ }cos\text{ }x\text{ }=\text{ }-2.5\]
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.