what are the values of x when x+2sincos=1?

Question

michele_laino · Answer

please, I don't understand your equation, please can you rewrite it?

anonymous · Answer

well the original equation was x+sin(x)^2 and I have to find where f prime is equal to one

michele_laino · Answer

Ok!, I write:
$$f(x)=x+(\sin x)^{2}$$
then
$$f'(x)=1+2 \sin x \cos x=1+\sin(2x)$$
from the condition:
$$f'(x)=1$$
we get:
$$1+\sin(2x)=1$$
so:
$$\sin(2x)=0$$
finally:
$$2x=k \pi$$
or:
$$x=\frac{ k \pi }{ 2 }$$
where:
$$k=0,\pm1,\pm2,\pm3,...$$

anonymous · Answer

where did the kpi come from?

michele_laino · Answer

generally the solution of equation:
$$\sin \alpha=0$$
are:
$$\alpha = k \pi$$
because in the unit circle, the points which have y-coordinate equal to zero are the only point located at 0°, 180°, 360°, and so on, whereas if we go in the negative sense, we have 0°, -180°, -360°, and so on
|dw:1416693193376:dw|