Pre-Calc help please!!!

1. The value of tan^-1(sqrt3) is pi/3. Write the general solution for x=arctan(sqrt3).

2. Explain why the range of y=cos^-1(x) is [0, pi] but the range of y=sin^-1(x) is [-pi/2, pi/2]

Question

Pre-Calc help please!!!

1. The value of tan^-1(sqrt3) is pi/3. Write the general solution for x=arctan(sqrt3).

2. Explain why the range of y=cos^-1(x) is [0, pi] but the range of y=sin^-1(x) is [-pi/2, pi/2]

anonymous · Answer

In part 1. you have to write the general solution for tan^-1(sqrt3)

anonymous · Answer

Yeah I know. Is it arctan(sart3) = pi/3 +/- 2pi(x)?

anonymous · Answer

Wait it would be pi(x) right...?

anonymous · Answer

Let me see

agent0smith · Answer

Don't use pi(x) use n or k or something, since x is already in the equation x=arctan(sqrt3).

anonymous · Answer

ya it is n(pi)+- pi/3

anonymous · Answer

Okay thanks. And what about the second problem?

anonymous · Answer

So what is the range of values of x where cos x is positive?

anonymous · Answer

Sorry my bad..  let me explain

anonymous · Answer

I need to draw

anonymous · Answer

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