Saved Bookmarks
| 1. |
If theta is positive acute angle then solve the equation:--4cos^2 theta -- 4 sin theta =1 |
| Answer» {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}\xa0is positive, therefore, taking{tex}\\sin \\theta = {1 \\over 2}{/tex}=> {tex}\\sin \\theta = \\sin {30^ \\circ }{/tex}=> {tex}\\theta = {30^ \\circ }{/tex} | |