What is the general solution to #cosx - tanx = 0#?

Question

2422 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-06T03:40:20

#x = 38.2˚ + 360˚n#

We have:

#cosx - tanx= 0#

#cosx - sinx/cosx = 0#

#(cos^2x - sinx)/cosx = 0#

#cos^2x - sinx =0#

#1- sin^2x - sinx = 0#

We let #u = sinx#.

#0 = u^2 + u - 1#

By the quadratic formula:

#u = (-1 +- sqrt(1^2 - 4 * 1 * -1))/(2 * 1)#

#u = (-1 +- sqrt(5))/2#

#sinx = (-1 +- sqrt(5))/2#

But our solutions will only be in the first quadrant, because we need cosine and tangent to have the same sign.

#x= arcsin((-1 + sqrt(5))/2)#

#x= 38.2˚#

This will only repeat after #360˚#, therefore,

#x = 38.2˚ + 360˚n#

Hopefully this helps!