Answers

2014-08-25T18:05:38+05:30
Consider the first part.

Tanx =  \sqrt{3}

The general solution is 

x =  \frac{\pi}{3} + 2\pi \theta and \frac{4\pi}{3}+2\pi\theta

where \theta is an integer.

Second part: Try yourself


1 5 1
The Brainliest Answer!
2014-08-25T18:48:17+05:30
tan x =  \sqrt{3}

Therefore
tan x = tan  \frac{ \pi }{3}
Therefore
x =  \frac{ \pi }{3}

Tangent is positive in I ^{st}III^{rd} Quadrant

Therefore general solution for tanx in I^{st} Quadrant is
x =  \frac{ \pi }{3}

And the general solution for tanx in III^{rd} Quadrant is
x = \pi +  \frac{ \pi }{3}
x =  \frac{4 \pi }{3}

And the principal solution for tanx is 
x = n \pi  +  \frac{ \pi }{3} , n∈Z

Thus in the similar way the general solution for cotx = - \sqrt{3} are 

x =  \frac{ 7\pi }{6}  \frac{13 \pi }{6}

And the principal solution of cotx 
x = n \pi +   \frac{ \pi }{3} , n∈Z

2 5 2
thnx fr ur help
Mntn nt...
:)
Plz mark as the best...