Question

Solve the following equation for all radian solutions and if
0 ≤ x < 2π.
Give all answers as exact values in radians. Do not use a calculator. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
2 sin2 x − sin x − 1 = 0

Answer 1

Assuming your 2sin2x is 2sin^2(x), this is a disguised quadratic. Let y=sin(x) then the equation becomes 2y^2-y-1=0 which factorises to (2y+1)(y-1)=0 giving y=-0.5 and y=1 and hence sin(x)=-0.5 and sin(x)=1. For the interval 0 ≤ x < 2π, there is only one solution for each of these:

sin(x)=1 => x=0.5π
sin(x)=-0.5 => x=1.5π

Hope this helps!

Answer 2

he is right^_^

Answer 3

but that last one should be 7pi/6 which is not 1.5pi

Answer 4

there is one more solution for the last one as well
where x=11pi/6 is a solution included in [0, 2pi]

Answer 5

Oops, thanks for the clarification, I did sin(x)=-1 by mistake.

Answer 6

thanks!