Question

F(x) =sinx and f(x)-f'(x) =0 then find x

Answer 1

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f(x) = sin x

so, f'(x) = d (sin x) / dx = cos x

now,

f(x) - f'(x) =0

$\implies \: sin \: x \: - \: \: cos \: x \: = 0 \\ \implies \: \: sin \: x \: = \: cos \: x \: \\ \implies \: \frac{sin \: x}{cos \: x} = 1 \\ \implies \: tan \: x \: = 1 \\ \implies \: tan \: x \: = \: tan \: \frac{\pi}{4} \\ \therefore \: \: x = \frac{\pi}{4} = {45}^{ \circ}$

Or,

we can write it in another way,

$tan \: x = \: tan \: \frac{\pi}{4} \\ \implies \: x = n\pi + \frac{\pi}{4} \\ \\ where \: \: \: \: n \in Z \:$

Cause we know that,

$tan \: \theta \: = tan \: \alpha \\ \implies \: \theta = n\pi + \alpha \\ where \: \\ n \in \: Z \:$