x = π/6 and π - π/6 --> x = π/6, 5π/6
Now solve sin(x) = 1...well there's only ONE place that sin(x) = 1 and that is where y = 1 on the unit @ 90° = π/2
This gives three solutions: x = π/6, π/2, 5π/6
4.) Same trick...
X² - X = 0
--> factor out X
X * (X - 1) = 0
--> plug back in cos(x)
cos(x) = 0
cos(x) - 1 = 0 --> cos(x) = +1
There are TWO places that cos(x) = 0 (two places where x = 0 on unit circle):
x = 90°, 270° = π/2, 3π/2
There is only ONE place where cos(x) = 1 (well technically there are two because you have [0, 2π]), so cos(x) = 1 when x = 1 on the unit circle which is @ x = 0° or x = 360° --> x = 0, 2π
So the solutions are x = 0, π/2, 3π/2 (and 2π if you really meant to include that)
Last problem...this is a little harder...but it's more or less the same.
5X + X - 2 = 0
--> I'm guessing you cannot factor, so just use the quadratic equation:
a = 5
b = 1
c = -2
x = (-1 ± √(1 - 4*5*(-2)) / (2*5) = (-1 ± √(21)) / 10 ~ -0.558257569, 0.358257569
--> solve
tan(x) = -0.558257569
tan(x) = 0.358257569
x = atan(-0.558257569) ~ -0.509160882
x = atan(0.358257569) ~ 0.344012204
So now we need to find the other two solutions (one of these in already in the 4th quadrant): -0.509160882 is equivalent to 2π - 0.509160882 ~ 5.77402443
For the first (negative) solution we need to find solutions in the 2nd and 4th quadrants:
2st quadrant: π - 0.509160882 ~ 2.63243177
4th quadrant: 2π - 0.509160882 ~ 5.77402443
For the second (positive) solution we need to find solution in 1st and 3rd quadrant:
1st quadrant: 0.344012204
3rd quadrant: π + 0.344012204 ~ 3.48560486
So the 4 solutions are:
x = 0.344012204, 2.63243177, 3.48560486, 5.77402443