Question

sin7 theta+sin 4 theta+sin theta=0 principal solution​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

We have,

$\sin(7 \theta) + \sin(4 \theta) + \sin( \theta) = 0$

$\implies\sin(7 \theta) + \sin( \theta) + \sin(4 \theta) = 0$

WE KNOW, $\rm \bold{sin(C)+sin(D)=2sin\bigg(\frac{C+D}{2} \bigg) cos\bigg(\frac{C-D}{2} \bigg)}$

So, applying this formula on first two terms,

$\implies2\sin \bigg( \frac{7 \theta + \theta }{2}\bigg)\cos \bigg( \frac{7 \theta - \theta }{2}\bigg) + \sin(4 \theta) = 0$

$\implies2\sin \bigg( \frac{8 \theta }{2}\bigg)\cos \bigg( \frac{ 6\theta }{2}\bigg) + \sin(4 \theta) = 0$

$\implies2\sin ( 4 \theta )\cos ( 3\theta ) + \sin(4 \theta) = 0$

$\implies\sin ( 4 \theta ) \{2\cos ( 3\theta ) + 1 \} = 0$

$\rm Either \: \: \sin ( 4 \theta ) = 0 \: \: \: or \: \: \: 2\cos ( 3\theta ) + 1 = 0$

$\rm \implies \: \: 4 \theta = n \pi \: \: \: or \: \: \: \cos ( 3\theta ) = - \frac{1}{2}$

$\rm \implies \: \: \theta = \frac{ n \pi}{4} \: \: \: or \: \: \: \cos ( 3\theta ) = \cos \bigg( \frac{2\pi}{3} \bigg )$

$\rm \implies \: \: \theta = \frac{ n \pi}{4} \: \: \: or \: \: \: 3\theta = 2m\pi \pm \frac{2\pi}{3}$

$\rm \implies \: \: \theta = \frac{ n \pi}{4} \: \: \: or \: \: \: \theta = \frac{2m\pi}{3} \pm \frac{2\pi}{9}$

Where, $\rm\:n,m\in\:I$