Math homework please help!!!!!?

Question

1. Find all solutions on [0,2pi] of ((secX^2)-1) - (cot(pi/2)-x) - 2 = 0

2. Find all solutions on [0,2pi] of 2sin(2X)cosX - 3sinX = 0

Answer 1

(1)

The Function is ((sec²(x) - 1) - cot(π/2 - x) - 2 = 0

tan²(x) - tan(x) - 2 = 0.

Let us Assume u = tan(x)

u² - u - 2 = 0.

Roots of the Function are

u² - 2u + u - 2 = 0

(u - 2)(u+1) = 0

Then u = 2 and -1

Now u = tan(x)

If u = 2

x = arctan(2)

x = 63.43 + kπ

If k = 0 then x = 63.43

If k = 1 then x = 243.43

If u = -1

x = arctan(-1)

x = -45 + kπ

If k = 1 then x = 135

Therefore Soltuion of (sec²(x) - 1) - cot(π/2 - x) - 2 = 0 in [0,2π] is 63.43, 135 and 243.43.

Answer 2

(2)

The Function is 2sin(2x)cos(x) - 3sin(x) = 0.

2(2sin(x)cos(x))cos(x) - 3sin(x) = 0

4cos²(x)sin(x) - 3 sin(x) = 0

sin(x) ( 4cos²(x) - 3) = 0

Therefore sin(x) = 0

then General soltuion of sin(x) = 0 then nπ +α for even multiples, (n+1)π + α for odd multiples.

If n = 0 then x = 0

4cos²(x) - 3 = 0 then  cosx = ±√3/2

If cos(x) = -√3/2

General soltuion of x = 2nπ + 5π/6

If n = 0 then x = 5π/6

If cos(x) = +√3/2

General soltuion of x = 2nπ + π/6

If  n = 0 then x = π/6

There solution of 2sin(2x)cos(x) - 3sin(x) = 0 is 0,π/6 and 5π/6.