How Do You Solve a Quadratic Trigonometric Equation?

AI Thread Summary
To solve the quadratic trigonometric equation 6sin²(x) - 6sin(x) + 1 = 0, one approach is to substitute sin(x) with a variable, such as y, transforming the equation into 6y² - 6y + 1 = 0. This quadratic can be solved using factoring or the quadratic formula, yielding up to two solutions for y. Each solution corresponds to sin(x), leading to two separate trigonometric equations to solve for x. Some participants suggest that it's acceptable to keep sin(x) in the equation rather than substituting it. Ultimately, the goal is to find x values within the interval [0, 2π].
ku07
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How do you algebraicly find the solution to this equation

6sin(to power of 2)x-6sinx+1=0

where 0<(or equal to) x <(or equal to) 2pi
 
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Let Sinx = u and then use the binomial theorem.
 
ku07 said:
How do you algebraicly find the solution to this equation

6sin(to power of 2)x-6sinx+1=0

where 0<(or equal to) x <(or equal to) 2pi
If you replace sin(x) by a variable y your equation will be :

6y^2 -6y + 1 = 0

Solve this equation for y. You will get maximum two solutions for y. each solution is equal to sin(x), so you will get two goniometric equations that will be easy to solve.

marlon
 
Im in trig as well and I am confused with some of yalls answers. Couldn't he treat this as a quadratic function. If it were me I would see if I could factor it then set each to 0 and solve; if it doesn't factor I would plug it in the quadratic formula and solve. I am in this same chapter of trig so please let me konw what I am overlooking.
 
seanistic said:
Im in trig as well and I am confused with some of yalls answers. Couldn't he treat this as a quadratic function. If it were me I would see if I could factor it then set each to 0 and solve; if it doesn't factor I would plug it in the quadratic formula and solve. I am in this same chapter of trig so please let me konw what I am overlooking.

yes you can treat it as a quadratic function, that's what marlon did in his post, although you could just leave sin (x) instead of replacing it with y, I believe that just plugging sin(x) = quadratic formula would work just as well, you would then have to do inverse sin on both sides to solve for x
 

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