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sinx-tanx=0
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Note that \[\large \begin{aligned}\sin x - \tan x = 0 &\implies \sin x - \dfrac{\sin x}{\cos x} = 0 \\ &\implies \sin x\left(1-\frac{1}{\cos x}\right) = 0\\ &\implies \sin x = 0\quad\text{ or }\quad 1-\frac{1}{\cos x}=0\\ &\implies\sin x = 0\quad\text{ or }\quad \cos x=1\end{aligned}\] Can you take things from here? :-)
sinx - tanx = 0 sinx - sinx/cosx = 0 (sinx cosx - sinx)/cosx = 0 set sinx cosx - sinx = 0 sinx (cosx - 1) = 0 sinx = 0 or cosx - 1 = 0 x = k(pi) where k is integer cosx = 1 x = turns out to be same solutions ANSWER: x = k(pi), k is an integer
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