Question

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

Answer 1

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

Explanation:

`cosx+sinxtanx = 2`

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

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

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

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

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

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

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

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`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`

Answer 2

`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`