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

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The discussion centers around solving the trigonometric equation 3cos(x) + sin(x) - 1 for positive values of x within the range of 0 to 360 degrees. The original poster struggled with the problem despite knowing the necessary equations and attempted to graph the function to find the x-intercept, but their results were incorrect. A suggested approach involves using the identity cos(x) = sqrt(1 - sin²(x)) to set up a quadratic equation in terms of cos(x) and then solving it while ensuring the solutions satisfy the Pythagorean identity. Another method proposed is to use parametric formulas with t = tan(x/2) to express sin(x) and cos(x) in terms of t, simplifying the equation further. The discussion highlights different strategies for tackling complex trigonometric equations effectively.
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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...
 
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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.
 
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.
 
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