Question

Total number of solutions of equation sin x tan 4x = cos x belonging to $(0,\pi )$ are:

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Answer 1

In this case also we have to simplify the given expression. For that we will write tan 4x = $\frac{sin4x}{cos4x}$

We do this to apply the formula for $sin(A\pm B)$ or $cos(A\pm B)$ after bringing the terms to one side.
sin4x $\frac{sin4x}{cos4x}$ = cosx
Multiplying by cos 4x on both the sides.

$\Rightarrow sinxsin4x=cosxcos4x(cos4x\ne 0)$

$\Rightarrow cos4xcosx-sinxsin4x=0$

$\Rightarrow cos(4x+x)=0$

$\Rightarrow cos5x=0$

$\Rightarrow$ 5x = (2n + 1)$\frac{\pi }{2}$
$\Rightarrow$ x = (2n + 1)$\frac{\pi }{10}$
Putting n = 1, 2, 3 we get,
x = $\frac{\pi }{10}$ ,$\frac{3\pi }{10}$ ,$\frac{5\pi }{10}$ ,$\frac{7\pi }{10}$ ,$\frac{9\pi }{10}$
the next value $\frac{11\pi }{10}$ — is out of our required range $(0,\pi )$.

$\Rightarrow$ we get 5 solutions.