InterviewSolution
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InterviewSolution
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The sum of all vaues of `theta in (0,(pi)/(2))` satisfying `sin^(2) x - sin 2x + sin 3x = 0`,isA. `(3pi)/(8)`B. `(5pi)/(4)`C. `(pi)/(2)`D. `pi` |
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Answer» Correct Answer - C Given, `sin^(2) 2 theta + cos ^(4) 2 theta = (3)/(4)` `rArr (1-cos^(2) 2 theta) + (cos^(4) 2 theta = (3)/(4) " "(because sin^(2) x = 1 - cos^(2) x)` `rArr 4 cos^(4) 2 theta - 4 cos^(2) theta +1= 0` `rArr ( 2 cos ^(2) 2theta - 1)^(2) = 0` `rArr 2 cos 2 theta -1 = 0` `rArr cos^(2) 2 theta = (1)/(2)` `rArr cos 2 theta = pm (1)/(sqrt(2))` If `theta in (0,(pi)/(2))`, then `2 theta in (0,pi)` `thereforecos 2 theta = (pi)/(sqrt(2))` `rArr 2 theta = (pi)/(4),(3pi)/(2), [ because cos((3pi)/(4))= cos (pi-(pi)/(4)) = - cos .(pi)/(4) = -(1)/(sqrt(2))]` `rArr theta = (pi)/(8),(3pi)/(8)` Sum of value of `theta= (pi)/(8) + (3pi)/(8) = (pi)/(2)` |
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