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If theta is positive acute angle …

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If theta is positive acute angle then solve the equation:--

4cos^2 theta -- 4 sin theta =1

  • 1 answers

Rashmi Bajpayee 6 years, 9 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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