How do you solve tana+1=0 and find all solutions in the interval [0,2pi)?

1 Answer
Oct 23, 2017

the solutions are -pi/4, (3pi)/4 and (7pi)/4

Explanation:

tanalpha+1=0rArrtanalpha=-1rArrtanalpha=-tan(pi/4)
it is of the form tanalpha=tank
the general solution is alpha=npi+krArralpha=npi-pi/4 since it is given that alphain[0,2pi) rArrn=0,1,2
on substituting n=0,1,2rArralpha=pi/4,(3pi)/4 and (7pi)/4