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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
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 :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.
The general form of a 2nd degree equation, also known as a quadratic equation, is ax^{2} + bx + c = 0, where a, b, and c are constants and x is the variable.
A 2nd degree equation can have a maximum of two solutions. These solutions can be real or complex numbers.
To solve a 2nd degree equation using the quadratic formula, first identify the values of a, b, and c in the equation ax^{2} + bx + c = 0. Then, substitute these values into the formula x = (-b ± √(b^{2} - 4ac)) / 2a to find the solutions for x.
Yes, a 2nd degree equation can have no solutions if the discriminant, b^{2} - 4ac, is negative. In this case, the solutions will be complex numbers.
2nd degree equations are used in various fields such as physics, engineering, and economics. They can be used to model and solve real-life problems such as calculating projectile motion, predicting the trajectory of a moving object, and analyzing financial data.