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Trignometry

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22 Jan 2012 15:36:33 IST

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Mathematics

if (cosx)^7 + (sinx)^4 = 1the find the value(s) of x, if it/they lie between pi to -pi

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kshitijaditya sharma

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22 Jan 2012 16:00:57 IST

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i found out the roots to be -pi/2 , 0 , pi/2

are there any other roots?

jagdish singh

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22 Jan 2012 18:59:41 IST

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$\hspace{-16}$Here $\mathbf{\cos^7x+\sin^4x=1}$ and $\mathbf{-\pi<x<\pi}$\\\\ $\mathbf{\cos^7x=1-\sin^4 x=(1-\sin^2 x).(1+\sin^2 x)}$\\\\ $\mathbf{\cos^7x=\cos^2 x.(1+\sin^2 x)\Leftrightarrow}$\\\\ $\mathbf{\cos^7x-\cos^2 x.(1+\sin x)=0}$\\\\ $\mathbf{\cos^2 x.\left\{\cos^5 x-(1+\sin^2 x)\right\}=0}$\\\\ So $\mathbf{\cos^2 x=0\Leftrightarrow x=\pm \frac{\pi}{2}}$\\\\ OR $\mathbf{\cos^5 x=1+\sin^2 x}$\\\\ Now $\mathbf{\cos^5 x\leq 1}$ and $\mathbf{1+\sin^2 x\geq 1\forall x\in \mathbb{R}}$\\\\ So only Possibilities is $\mathbf{\cos^5 x =1\Leftrightarrow x=0}$\\\\ and $\mathbf{1+\sin^2x=1\Leftrightarrow \sin^2 x=0\Leftrightarrow x=0}$\\\\ bcz Here $\mathbf{-\pi<x<\pi}$\\\\ So $\boxed{\boxed{\mathbf{x=0\;,\pm \frac{\pi}{2}}}}$$

Manasi

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25 Jan 2012 00:26:05 IST

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Now, either cos² x = 0 or cos⁵ x = 1+sin² x

Case 1: cos² x = 0

=> x = {-pi/2 , +pi/2}

Case2: cos⁵ x = 1+sin² x .............................(A)

Now here's the logic part.

For equation A to have a solution, i.e. for both the sides to be equal, thr respectives ranges shud intersect/overlap each other, at least at one point.

Range of cos⁵ x= [-1,1]

Range of sin² x= [0,1] (since, its a square term)

=> Range of (1+ sin² x) = [1,2]

Thus, it can be clearly seen, that there is only one point of intersection, which is when both

cos⁵ x =1= 1+sin² x

=> x = 0

Thus the solution set is x = {-pi/2, 0, pi/2}

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