If theta is positive acute angle …
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Posted by Abhishek Kapoor 6 years, 10 months ago
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Rashmi Bajpayee 6 years, 10 months ago
{tex}4{\cos ^2}\theta - 4\sin \theta = 1{/tex}
=> {tex}4\left( {1 - {{\sin }^2}\theta } \right) - 4\sin \theta = 1{/tex}
=> {tex}4 - 4{\sin ^2}\theta - 4\sin \theta = 1{/tex}
=> {tex} - 4{\sin ^2}\theta - 4\sin \theta + 4 - 1 = 0{/tex}
=> {tex}4{\sin ^2}\theta + 4\sin \theta - 3 = 0{/tex}
=> {tex}4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0{/tex}
=> {tex}2\sin \theta \left( {2\sin \theta + 3} \right) - 1\left( {2\sin \theta + 3} \right) = 0{/tex}
=> {tex}\left( {2\sin \theta + 3} \right)\left( {2\sin \theta - 1} \right) = 0{/tex}
=> {tex}2\sin \theta + 3 = 0{/tex} or {tex}2\sin \theta - 1 = 0{/tex}
=> {tex}\sin \theta = {{ - 3} \over 2}{/tex} or {tex}\sin \theta = {1 \over 2}{/tex}
But {tex}\theta {/tex} is positive, therefore, taking
{tex}\sin \theta = {1 \over 2}{/tex}
=> {tex}\sin \theta = \sin {30^ \circ }{/tex}
=> {tex}\theta = {30^ \circ }{/tex}
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