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root 2 cosx+ root6 sinx= max (sintheta + costheta) where theta belongs to R find the general value of x

root 2 cosx+ root6 sinx= max (sintheta + costheta) where theta belongs to R find the general value of  x

Grade:11

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:Hello student, please find answer to your question
\sqrt{2}cosx + \sqrt{6}sinx = max(sin\theta +cos\theta )
f(\theta ) = \sin\theta +cos\theta
f'(\theta ) = cos\theta -sin\theta = 0
\theta = n\pi + \frac{\pi }{4}
[f(\theta )]_{max} = \sqrt{2}
\sqrt{2}cosx + \sqrt{6}sinx = \sqrt{2}
cosx + \sqrt{3}sinx = 1
\frac{1}{2}.cosx + \frac{\sqrt{3}}{2}sinx = \frac{1}{2}
sin(x + \frac{\pi }{6}) = sinx.\frac{\sqrt{3}}{2} + cosx.\frac{1}{2}
sin(x + \frac{\pi }{6}) = \frac{1}{2}
x + \frac{\pi }{6} = 2n\pi + \frac{\pi }{6}
x = 2n\pi
keerthi priya
35 Points
9 years ago
thnk u very much

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