Solve SinA- tanA=0?

3 Answers
Apr 26, 2018

#" The Solution Set"={kpi : k in ZZ}#.

Explanation:

#sinA-tanA=0#.

#:. sinA-sinA/cosA=0#.

#:. (sinAcosA-sinA)/cosA=0#.

#:. sinAcosA-sinA=0#.

#:. sinA(cosA-1)=0#.

#:. sinA=0, or, cosA=1#.

But, #cosA=1 rArr sin^2A=1-cos^2A=1-1^2=0, i.e., #

#cosA=1 rArr sinA=0#.

#:. {A : cosA=1} sub {A : sinA=0}#.

Hence, we need to consider #sinA=0# only.

But, #sinA=0 rArr A=kpi, k in ZZ#.

#:." The Solution Set"={kpi : k in ZZ}#.

Apr 26, 2018

The solution is #A={kpi}#, #AA k in ZZ#

Explanation:

The equation is

#sinA-tanA=sinA-sinA/cosA=0#

#sinA(1-1/cosA)=0#

Therefore,

#{(sinA=0),(1-1/cosA=0):}#

#<=>#, #{(sinA=0),(cosA=1):}#

#<=>#, #{(A=kpi),(A=2kpi):}#, # AA k in ZZ#

The solution is

#A={kpi}#, #AA k in ZZ#

Apr 26, 2018

#A = pin#

Explanation:

The first step is to find a way where we can have only have one trig function in the equation. Remember that cosine and secant are reciprocals (can be readily written as one or the other). Now you can take advantage of pythagorean identities: #sin^2theta + cos^2theta = 1#.

#sqrt(1 - cos^2A) = tanA#

Square both sides:

#1 - cos^2A = tan^2A#

Now recall that #tan^2A + 1 =sec^2A#. If you use this you will only be left with cosines and secants which was our goal.

#1- cos^2A =sec^2A - 1#

#2 = 1/cos^2A + cos^2A#

#2 = (1 + cos^4A)/cos^2A#

#2cos^2A = cos^4A + 1#

#0 = cos^4A - 2cos^2A + 1#

This is a simple perfect square trinomial.

#0 = (cos^2A - 1)^2#

#cosA = +- 1#

#A = 0 or pi-> A = pin#

Hopefully this helps!