My problem reads
"Solve 6sin2x + sinx - 2 for all x in 0 < x < pi/2"
I know through graphing that the answer is pi/6. Is there another way of proving this?
You may also treat it as a quadratic equation and then the values can be found. For ax^2 + bx + c = 0; we have x = [ -1 +/- sqrt ( b^2- 4ac )]/ 2a
In this case you will get sin x = [ -1 +/- sqrt ( 1+ 48 )]/ 12
now solve it further