Barron Page 77 Solve for 2sinx+ cos2x=2sin^2x-1 for 0< and equals x < and equals 2$\displaystyle \pi$? Thanks in advance! Urgent!!
Follow Math Help Forum on Facebook and Google+
Hint: $\displaystyle \cos 2x = 1 - 2\sin^2 x$ Arrange your equation and you will have it in the form of: $\displaystyle ay^2 + by + c = 0$ Then you can use the quadratic formula.
Ok thanks. I got 4sin^2x-sinx+2=0 do i just solve x? I don't have to consider sin?
You solve the quadratic taking sin(x) as the variable and then you find x with x = arcsin(sin(x)) there can be 2 solutions.
How do you when to use arcsin or when to use sin?
Suppose $\displaystyle a\cdot sin(x)^2 + b \cdot sin(x) + c = 0 $ $\displaystyle sin(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ $\displaystyle x = arcsin(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) $
View Tag Cloud