Question

solve for x: sinx + sin3x + sin5x = 0

Answer 1

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sin x + sin 3x + sin 5x = 0
(sin x + sin 5x) + sin 3x = 0

$[2 sin (\frac{x+5x}{2}). cos (\frac{x-5x}{2} )] + sin 3x = 0$

$(2 sin \frac{6x}{2}. cos \frac{-4x}{2}) + sin 3x = 0$

$(2 sin 3x. cos(-2x)) + sin 3x = 0$

$(sin 3x. cos 2x)+ sin 3x = 0$

$sin 3x(2cos 2x + 1) = 0$

Hence,
Sin 3x = 0 -----(1)

2 cos 2x + 1 = 0
=> 2 cos 2x = -1
=> $cos 2x = \frac{-1}{2}$ ----(2)

Now,
Find general solution of (1) & (2) separately

¶¶¶ General Solution of sin 3x = 0

sin 3x = 0

¶sin kx = 0 => kx = nΠ
=> $x = \frac{nΠ}{k}$ where n£Z

•°• 3x = nΠ => x = $\frac{nΠ}{3}$
where n£Z

¶¶¶ General solution of Cos 2x =$\frac{-1}{2}$

Let cos x = cos y
=> cos 2x = cos 2y --------(3)

Given,
cos 2x = $\frac{-1}{2}$ ------(4)

From (3)&(4)
cos 2y = $\frac{-1}{2}$

cos 2y =cos $\frac{2Π}{3}$

=> 2y = $\frac{2Π}{3}$

General solution for cos 2x = cos 2y is
2x = 2nΠ (+-) 2y where n£Z

=> 2x = 2nΠ (+-) $\frac{2Π}{3}$

=> x = $\frac{1}{2}$(2nΠ (+-) $\frac{2Π}{3})$

=> x = [$\frac{2nΠ}{2}$ (+-) $\frac{1}{2}] × \frac{Π}{3}$

=> x = nΠ (+-) $\frac{Π}{3}$
where n£Z

Hence,
General Solution is
For sin x = 0 , x =$\frac{nΠ}{3}$
For cos 2x = $\frac{-1}{2}$, x = nΠ (+-) $\frac{Π}{3}$

where n£Z

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¶¶¶ Identities Used :

•
$sin C + sin D = 2 sin (\frac{C + D}{2}) cos (\frac{C + D}{2})$

• $cos(-x) = cos x$

^^ Also note :
$cos \frac{-1}{2} = cos \frac{2Π}{3}$

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