Question

Question #fb399

Answer 1

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.

Answer 2

`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`