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

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SUMMARY

The discussion centers on solving the trigonometric equation 3cos(x) + sin(x) - 1 for x in the range [0, 360] degrees. A user initially attempted to graph the equation to find the x-intercept but encountered incorrect values. The solution involves transforming the equation into a quadratic form using the identity cos(x) = sqrt(1 - sin²(x)) and applying parametric formulas with t = tan(x/2) to express sin(x) and cos(x). This method simplifies the problem and ensures valid solutions are found.

PREREQUISITES
  • Understanding of trigonometric identities, specifically cos(x) and sin(x).
  • Familiarity with quadratic equations and their solutions.
  • Knowledge of parametric equations and their applications in trigonometry.
  • Graphing techniques for visualizing trigonometric functions.
NEXT STEPS
  • Study the derivation and application of the parametric formulas for sine and cosine.
  • Learn how to solve quadratic equations derived from trigonometric identities.
  • Explore advanced graphing techniques for trigonometric functions to identify intercepts accurately.
  • Investigate the implications of the unit circle in solving trigonometric equations.
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Students preparing for mathematics exams, educators teaching trigonometry, and anyone interested in solving complex trigonometric equations.

wasteofo2
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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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