What are the solutions on #0 ≤ x < 2pi# to #secx + cscx + 1/tanx - tanx = 0#?

#1/cosx + 1/sinx + 1/(sinx/cosx) - sinx/cosx= 0#

#1/cosx + 1/sinx + cosx/sinx - sinx/cosx= 0#

#sinx + cosx + cos^2x- sin^2x = 0#

#sinx +cosx = sin^2x- cos^2x#

#sinx + cosx = (sinx - cosx)(sinx + cosx)#

#0 = (sinx - cosx)(sinx + cosx) - (sinx + cosx)#

#0 = (sinx + cosx)(sinx - cosx - 1)#

Solving #sinx + cosx = 0#

#sin^2x + cos^2x + 2sinxcosx = 0#

#1 + 2sinxcosx = 0#

#sin(2x) = -1#

#2x= (3pi)/2#

#x = (3pi)/4#

But this is only one solution in #0 ≤ x ≤ 2pi#. We note that to have a valid solution, #sinx = -cosx#. This occurs in quadrants #I# and #IV#. So the other solution for the first equation is #(7pi)/4#.

Solving #sinx - cosx = 1#

#sin^2x + cos^2x- 2sinxcosx = 1#

#-2sinxcosx+1 = 1#

#-sin(2x) = 0#

#sin(2x) = 0#

#2x = 0, pi#

#x = 0, pi/2#

We also note that since #(2pi)/2 = pi#, we should have another solution in the domain. But this is extraneous, because #cos(pi) = -1# so #sinx + cosx = -1 != 1#.

We know that #cotx = 1/tanx#, which we had in the initial equation. The cotangent function is undefined when #sinx= 0#, or when #x = 0#. Since we had tangents in the initial equation, we can say that #cosx != 0#, so #x!= pi/2#.

This means our two solutions are in fact #x= (3pi)/4# and #x= (7pi)/4#.

#1/(cos x) + 1/(sin x) + cos x/(sin x) - sin x/(cos x) = 0# Take (sin x.cos x) as common denominator --> Condition (1): #sin x.cos x != 0#, or #x != pi/2# and #x != pi#

#sin x + cos x + cos^2 x - sin^2 x = 0# #sin x + cos x + (cos x - sin x)(cos x + sin x) = 0# #(sin x + cos x)(1 + cos x - sin x) = 0# Either one of the factors should be zero a. sin + cos x = 0 Use trig identity: #sin x + cos x = sqrt2cos (x - pi/4)# #sin x + cos x = sqrt2cos (x - pi/4) = 0# cos (x - pi/4) = 0 --> 2 solutions #x - pi/4 = pi/2# --> #x = pi/2 + pi/4 = (3pi)/4# #x - pi/4 = (3pi)/2# --> #x = (3pi)/2 + pi/4 = (7pi)/4# b. 1 + cos x - sin x = 0 sin x - cos x = 1 Use trig identity: #sin x - cos x = - sqrt2cos (x + pi/4)# --> #sin x - cos x = - sqrt2cos (x + pi/4) = 1# #cos (x + pi/4) = - 1/sqrt2 = - sqrt2/2# --> 2 solutions #x + pi/4 = +- (3pi)/4# #x + pi/4 = (3pi)/4# --> #x = (3pi)/4 - pi/4 = pi/2# #x + pi/4 = - (3pi)/4# --> #x = - (3pi)/4 - pi/4 = - pi# or #x = pi# These values #pi/2 and pi# aren't the answers because of the above condition (1).