Trigonometric Equations: Please Help!?

Question

I am tearing my hair out here trying to figure these problems out. I would really appreciate it if someone could show me the steps they took to figure these out. Thanks much.

1) Find all solutions of tanX - cotX = 0, where is X is measured in degrees and 0
2) Find all solutions of cscX = -3, where X is measured in degrees and 0
3) Find all exact solutions of 2cos^2x-sinx = 1 in the interval 0

I know there's a couple of 'em, but I'm hoping with some explanation I can use these three as a basis for understanding how to more. Please help me with this. Thank you very much!

Answer 1

1) Find all solutions of tanX - cotX = 0

Same as writing

tanX = cotX
tanX = 1/(tanX)
tan^2X = 1
tanX = +/- 1

tanX is positive or negative 1 in four places (45, 135, 225, and 315 degrees), between 0 and 360 degrees.

2) cscX = -3

I find it easier to work with its reciprocal:

sinX = -1/3
X = ~199.471 degrees, ~340.529 degrees, between 0 and 360.

3) 2cos^2x-sinx = 1

cos^2x = (1 - sin^2x), so this is:

2(1 - sin^2x) - sinx = 1
2 - 2sin^2x - sinx = 1
2sin^2x + sinx - 1 = 0

Let u=sinx:

2u^2 + u - 1 = 0
u = [ -1 +/- sqrt(1^2 - 4*2*(-1)) ] / ( 2*2)
u = [ -1 +/- sqrt(1 + 8) ] / 4
u = [ -1 +/- sqrt(9) ] / 4
u = [ -1 +/- 3 ] / 4
u = -1, or u= 1/2

Two solutions for sinx, treated separately:

sinx = -1
x = 3pi/2

sinx = 1/2
x = pi/6, 5pi/6

The solutions between 0 and 2pi are:
pi/6, 5pi/6, and 3pi/2.

Answer 2

1)
tanX - cotX = 0
sinX/cosX = cosX/sinX
sin^2X = cos^2X
sin^2X = 1 - sin^2X
sin^2X + sin^2X = 1
2 sin^2X = 1
sin^2X = 1/2
sinX = sqrt(1/2)
sinX = +- sqrt(2) / 2
x = 45 , 135 , 225 , 315

2)
cscX = -3
1/sinX = -3
sinX = -1/3

3)
2cos^2x-sinx = 1
2(1-sin^2X) - sinX = 1
2 - 2sin^2X - sinX = 1
-2sin^2X - sinX + 1 = 0
-2sin^2x -2sinX + sinx + 1 =0
-2sinX(sinX+1) + (sinX+1) = 0
(1-2sinX)(sinX+1) = 0
sinX = 1/2
x = 30 , 150
or
sinX = -1
x = 270