Find the solution

Question

a) sin(x - 2) = 1

b) 2sin(theta) - 1 = 0

c) sin(theta) = tan(theta)

d) sin(theta) * cos(theta) = 0

e) sin^ 2 (x)-sin(x)=0,x in[0,2 pi[

f) cos( theta+ pi/2)=sin( theta), theta in[0,2 pi[

Answer 1

a) x-2 = 2npi/2 where n=any integer, so x = 2+2npi/2

unless x is measured in degrees, then x=2+n90 where n= any integer

such as 92, -268, 352 degrees

b) 2sin(theta) = 1, sin(theta) = 1/2, theta = 2npi/3 or 4npi/3 or the equivalent in degrees

theta = 60 + n360 or 120 + n360 where n= any integer

c) sin(theta) = tan(theta)

theta = 0 or pi

sin(theta) = tan(theta) = sin(theta)/cos(theta)

cos(theta)sin(theta) - sin(theta) =0

sin(theta)[cos(theta) -1] = 0

sin(theta) = 0, cos(theta) = 1

theta = 0 or pi, theta = 0

do the plus 2pin to each solution where n= any integer

d) sin(theta)cos(theta) = 0

set each factor =0 and solve

sin(theta) = 0 or cos(theta) = 0

theta = 0 or pi, or theta = pi/2

each with plus 2pin where n = any integer

e) sin^2x - sinx =0

factor out sinx

sinx(sinx-1) =0 set each factor =0

sinx = 0 and sinx=1

x = 0 or pi and x=pi/2

each solution with plus 2pin where n = any integer

f) cos(theta + pi/2) = sin(theta)

theta = 0 or pi, maybe 2pi also, but hard to tell if you included 2pi in the interval of theta