हल कीजिए- ` sin ^(2 ) theta + 2 cos theta + 1/4 = 0.`

Explanation

` sin^(2) theta + 2 cos theta + 1/4 = 0 `
` rArr 4 sin^(2) theta + 8 cos theta +1 = 0 `
` rArr 4 (1-cos^(2)theta) + 8 cos theta +1=0`
` rArr 4 cos^(2) theta - 8 cos theta - 5 = 0 `
` rArr 4 cos^(2) theta - 10 cos theta + 2 cos theta - 5 = 0 `
` rArr 2 cos theta ( 2 cos theta - 5) +1 (2 cos theta - 5) = 0 `
` rArr ( 2 cos theta - 5) (2 cos theta+1)= 0 `
` rArr 2 cos theta - 5 = 0 " या " 2 cos theta +1= 0 `
अब ` 2 cos theta - 5 = 0 `
` rArr cos theta = 5/2 ` जो संभव नहीं है ।
और ` 2 cos theta +1 = 0 `
` rArr cos theta = -1/2 = - cos. pi/3`
` os (pi - pi/3) = cos. (2pi)/3 `
` :. theta = 2 n pi pm (2pi)/3 `