Solving quadratic trig equations

fred2028

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Apr 10, 2006
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In our functions course, the government has removed several parts, including solving quadratic trig equations. So I was wondering, what does a quadratic trig equation look like, and in what ways can it be solved?
 
fred2028 said:
In our functions course, the government has removed several parts, including solving quadratic trig equations. So I was wondering, what does a quadratic trig equation look like, and in what ways can it be solved?

A quadratic equation is of the form \(\displaystyle ax^2 + bx + c = 0\) where a,b,c are real numbers.

A quadratic trigonometric equation is of the form \(\displaystyle a[f(x)]^2 + b[f(x)] + c = 0\) where f(x) is the sine, cosine, tangent or any of their reciprocle functions and a,b,c are real numbers.

For example:

\(\displaystyle cos^2(x) - 2cos(x) + 1 = 0\)

Can be solved just as \(\displaystyle x^2 -2x + 1=0\) can be solved. The only difference is when you factor and solve you need to take the inverse cosine to solve for x:

\(\displaystyle (cos(x) - 1)^2 = 0 \\ cos(x) = 1 \\ x = cos^{-1}(1) \\ x = 0, \,\, 2\pi, \,\, 4\pi...\) etc.
 
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