The discussion focuses on solving the quadratic trigonometric equation 6sin²(x) - 6sin(x) + 1 = 0 for the interval 0 ≤ x ≤ 2π. Participants suggest substituting sin(x) with a variable y, transforming the equation into 6y² - 6y + 1 = 0, which can be solved using the quadratic formula. The solutions for y correspond to the values of sin(x), leading to two goniometric equations that can be solved for x. The conversation emphasizes that treating the equation as a quadratic function is valid and effective.
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If you replace sin(x) by a variable y your equation will be :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
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.