\displaystyle{x}\approx{36.869} ... Explanation: You need to rearrange the equation: \displaystyle{\sin{{x}}}=\frac{{3}}{{5}} \displaystyle{x}={{\sin}^{{-{1}}}{\left(\frac{{3}}{{5}}\right)}}\approx{36.869}\ldots

This is difficult to prove since it is not true Explanation: As a simple counter example, consider the case when \displaystyle{\left(\text{XXX}\right)}{x}={0} \displaystyle{\sin{{\left({2}{\left({0}\right)}\right)}}}={0}\ne{1}

It is well-known that over a full period, -1\le\sin x\le1 and all values in this range can be reached. Then -3\le-3\sin x\le3 and 1\le4-3\sin x\le7. Starting from the standard sinusoid, ...

I would suggest that the most critical information here is dx/dt = c as it shows how constant values propagate. So using this fact we have: \frac{dx}{dt} = c \rightarrow x = ct + x_0 Which ...

You are forgetting to multiply by u'. f'(x) = \cos(u) \cdot u'. To compute the derivative of u = \ln(\cos(x^2+1)) We must use the chain rule once again. Now let w = \cos(x^2+1) and keep ...

Note that \sin(x) increases from x = 0 to x = {\pi \over 2}, then decreases from {\pi \over 2} to \pi, in a way symmetric about {\pi \over 2}. So for a given 0 \leq \alpha \leq 1, the x \in [0,\pi] ...