Solve the following. Find all solutions from 0 degrees to 360 degrees.

Question

1) 4 sin² x – 1 = 0

2) 2 cos² x = 1

3) 2 sin² x + sin x = 1

4) cos² x + 2cos x = 3

5) 2 sin² x + 7 sin x = 4

6) 2 sin² x – 1 = 0

7) cos x sin x + sin x = 0

8) sin x cos x – cos x = 0

9) 2 sin² x - 1 = 0

10) sin² x + 2 sin x – 3 = 0

show all work please. thank you.

Answer 1

4sin^2(x) = 1

sin^2(x) = 1/4

sin(x) = + or - 1/2

use an inverse trig function on a calculator if you don't recognize the 30-60-90 right triangle

x= 30 or 150 degrees

2cos^2(x) = 1

cos^2(x) = 1/2

cos(x) = - or + sqr2/2

x = 45 or 135 degrees

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

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

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

sin270=-1, sin30=1/2 sin(180-30)=1/2

x=270, 30, 150 degrees

cos^2(x) +2cos(x) -3 = 0

(cos(x)+3)(cos(x)-1) = 0

cos(x)=-3 or cos(x)= 1

x = 0 degrees

(ignore cos(x)=-3 which is impossible. as -1<or =cos(x)<or =1 for any angle x)

2sin^2(x)+7sin(x) -4 = 0

(2sin-1)(sin(x)+4) = 0

sin(x)= 1/2

x = 30, 150 degrees

2sin^2(x) = 1

sin^2(x) = 1/2

sin(x) = + or - sqr2/2

x = 45 or 135 degrees

cos(x)sin(x) + sin(x) = 0

sin(x)(cos(x)+1) = 0

sinx=0 or cos(x) =-1

x = 0, 180 degrees

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

cos(x)(sin(x)-1) = 0

cos(x) =0 or sin(x) =1

x = 90, 270 degrees

2sin^2(x) - 1 =0

sin^2(x) =1/2

sin(x) = - or +sqr2/2

x = 45 or 135 degrees

sin^2(x) +2sin(x) -3 = 0

(sin(x)+3)(sin(x)-1) = 0

sin(x) =1 (ignore sin(x)=-3 as it's impossible)

x= 90 degrees