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

2 Answers
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

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