How do you solve #cos x + sin x tan x = 2# over the interval 0 to 2pi?

2 Answers
Ahmemaru
May 9, 2018

#x = pi/3#
#x = (5pi)/3#

Explanation:

#cosx+sinxtanx = 2#

#color(red)(tanx = (sinx)/(cosx))#

#cosx+sinx(sinx/cosx) = 2#

#cosx+sin^2x/cosx = 2#

#cos^2x/cosx+sin^2x/cosx = 2#

#(cos^2x+sin^2x)/cosx = 2#

#color(red)(cos^2x+sin^2x=1)#

#color(red)("the phythagrean identity")#

#1/cosx = 2#

multiply both sides by #cosx#

#1 = 2cosx#

divide both sides by #2#

#1/2 = cosx#

#cosx = 1/2#

from the unit circle #cos(pi/3)# equals #1/2#

so

#x = pi/3#

and we know that #cos# is positive in the first and fourth quadrant so find an angle in the fourth quadrant that #pi/3# is the reference angle of it

so

#2pi - pi/3 = (5pi)/3#

so

#x = pi/3, (5pi)/3#

Dean R.
May 9, 2018

#x = pi/3 or {5pi}/3 #

Explanation:

The way I'm checking the other answer is writing my own.

#cos x + sin x tan x = 2#

# cos x + sin x (sin x / cos x) = 2 #

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

# 1 = 2 cos x #

# cos x = 1/2 #

There's the cliche triangle, you knew it was coming.

In the range,

#x = pi/3 or {5pi}/3 #

Check:

# cos({5pi}/3) + sin ({5pi}/3) tan({5pi}/3) = 1/2 + -\sqrt{3}/2 cdot {-sqrt{3}//2}/{1//2} = 1/2 + 3/2 = 2 quad sqrt #

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