PRECALCULUS PROBLEM WITH SOLVING FOR X

Please help, 10 pts best answer!!!!!!!!

Solve for x in the interval [0,2pi], And express each angle in terms of pi (if possible) or round to the nearest ten-thousandth.

1. tan^2 x-1=0

2. 2sin^2 x-3 sin x+1=0

4. cos^2 x- cos x=0

PLEASE EXPLAIN!

also, using the quadratic formula, solve for x to the nearest ten thousandth in the interval [0,2pi]:

1. 5tan^2 x+ tanx-2=0

I'll believe the 10pts when I see it...

1.)

tan²(x) - 1 = 0
--> what you have looks like this:

X² - 1 = 0
--> factor (you have a difference of squares)

(X + 1)(X - 1) = 0
--> set each factor to 0

X + 1 = 0 --> X = -1
X - 1 = 0 --> X = +1
--> but remember X = tan(x)

so we really have:
tan(x) = +1
tan(x) = -1

Find the solutions...well tan(x) = 1 @ 45°, which is π/4 radians. But this is just the 1st quadrant result.

The tangent is positive when either BOTH cos and sin are positive, or BOTH are negative. cos and sin are BOTH positive in the 1st quadrant (i.e. π/4) and cos and sin are BOTH negative in the 3rd quadrant: π + π/4 = 5π/4. tangent is negative when either cos or sin is negative, but NOT both. So tangent is negative in the 2nd and 4th quadrants: π - π/4 and 2π - π/4
-->

tan(x) = +1 --> x = π/4, 5π/4
tan(x) = -1 --> x = 3π/4, 7π/4

So the 4 solutions are: x = π/4, 3π/4, 5π/4, 7π/4

2.)
Same trick, substitute X = sin(x)
-->

2X² - 3X + 1 = 0
--> factor

(2X - 1)(X - 1) = 0
--> set each factor to 0 and solve for X

2X - 1 = 0 --> X = 1/2
X - 1 = 0 --> X = 1
--> put the sin's back in

sin(x) = 1/2
sin(x) = 1

1/2 is from our 30-60-90 triangle. 2 is the hypotenuse and 1 is the smaller side (the other is √3). The smaller angle for sin means sin(x) = 1/2 --> x = 30° = π/6.

So where is sin positive? sin is positive when the y coordinates are positive, this means quadrant 1 and quadrant 2: