What is the solution to the equation cosx=tanx for -180<= x <= 180?

I can only use the identities, sin^2x + cos^2x =1 and tanx = sinx/cosx, and of course factorisation. I am getting approx. values of 38 and 142 degrees.

Answers:
cosx=tanx
cosx=sinx/cosx
cos^2(x)=sinx
1-sin^2(x)=sinx {pythagoras}
sin^2(x)+sinx-1=0
use quadratic formula
sinx=(-1+or-sqrt(5))/2
=0.6180339888
or -1.6180339888
sine has to lie between +/-1
sinx= 0.6180339888
x=arcsin(0.6180339888)
=38.17270763
+ve value for sine lies in 1st and 2nd
quadrants
therefore,x=38.17270763 degrees or
141.8272924 degrees

these two values were checked in the
original equation and were found
to be correct

i hope that this helps
X= -1/2

OR

X= -0.5
You are absolutely right.
sin x =-0.5 +/- (root5)/2
So x =38.2deg. or 141.8deg. (to 2dp)
cos(x)=sin(x)/cos(x)
(cos(x))^2- sin(x) =0
1 - (sin(x))^2 - sin(x) = 0
(sin(x))^2 + sin(x) - 1 =0

Sin(x)=0.61,-1.61

Sin value ranges between -1 to +1 so -1.61 is not valid

x= 38.17 Degrees
cosx = sinx/cosx --->( cosx) ^2 = sinx = 1-sin(x)^2

this is equivalent to (sinx)^2 +sinx -1 =0

sinx =( -1+ (1+4)^0.5)/2 = 0.618 2solutions 38° or 142 °

The negative solution gives sinx <-1 impossible

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