Question #fb399

2 Answers
Dec 7, 2017

Assuming #sin^2(x)=3cos^2(x)#, then first let's change it in to #tan^2(x)#,

#sin^2(x)/cos^2(x) = 3(cos^2(x)/cos^2(x))# #rarr# #tan^2(x) = 3#

Now from here all we need to do is take the square root of both sides,

#sqrt(tan^2(x)) = +-sqrt(3)# #rarr# #tan(x) = +-sqrt(3)#

Remember that because we will be taking the inverse tangent we will need to apply the proper restrictions on its domain and range. #tan^-1(x); D:(-oo, oo) ;R:( pi/2,-pi/2)#.

Knowing this we can gather that the answers for when #tan(x) = +-sqrt(3)#, are,#pi/3# and #5pi/3# uising the unit circle or right triangle method.

Dec 7, 2017

#pi/3; (2pi)/3; (4pi)/3; (5pi)/3#

Explanation:

#sin^2 x - 3cos^2 x = 0#
#(sin x - sqrt3cos x)(sin x + sqrt3cosx) = 0#
Either one of the factor should be zero
a. #sin x - sqrt3cos x = 0#.
Divide both sides by cos x (condition #cos x != 0#)
#tan x - sqrt3 = 0# --> #tan x = sqrt3#
Trig table and unit circle give -->
#x = pi/3# and #x = p/3 + pi = (4pi)/3#
b. #sin x + sqrt3cos x = 0#
Divide both sides by cos x.
#tan x = - sqrt3#
#x = (2pi)/3#, and #x = (2pi)/3 + pi = (5pi)/3#