The discussion revolves around solving a quadratic trigonometric equation of the form 6sin²(x) - 6sin(x) + 1 = 0, with the variable x constrained between 0 and 2π. Participants explore various algebraic approaches to find solutions.
There is an active exploration of different methods to approach the problem, with participants sharing their thoughts on substitution and the use of the quadratic formula. Some express confusion about the various suggestions and seek clarification on the reasoning behind them.
Participants note that the equation is part of a trigonometry chapter, indicating a shared context of learning and understanding quadratic functions within trigonometric scenarios.
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.