Solving the Tricky 3cos(x) + sin(x) - 1 Problem

[wasteofo2]

On my math final, there was this bastard of a trig problem that I simply couldn't solve. I knew the equations to use, how to solve the problem, but the answers just didn't work...

Anyway, the question was this:
Find all positive values of x for x being greater than or equal to zero, and less than or equal to 360.
3cos(x) + sin(x) - 1

How would you go about solving this? I tried graphing the equation and finding the x-intercept, but the values i got didn't work for whatever reason...

 

[mathman]

When you say 360, do you mean degrees?
Since cos(x)=sqrt(1-sin²(x)), you can set it up as a quadratic equation in cos(x), solve and go from there - discarding any solutions where sin²+cos² does not add up to 1.

 

[Maxos]

The best way when you have a linear trig. equation is to use parametric formulae

t=tan(x/2)

then sin(x)=2t/(t^2+1) and cos(x)=(1-t^2)/(1+t^2)

you substitute them into the equation and solve it into t, simple, isn't it?

Never heard about that? Quite strange.