Solve the trigonometric equation Sin x 0493 for 0 degree l
Solution
a. Sin x=-.493
Taking sin-1 on both sides
sin-1(sin x)=sin-1-0.493
x=-29.54degree,209.54 degree,330.46 degree
and x have to be between 0 and 360
Therefore x=209.54 degree and 330.46 degree
b. 3 cos^2theta - 2 sin theta-2=0
cos^2 theta=1-sin^2 theta
3(1-sin^2 theta)-2sin theta -2=0
Let sin theta=t
3(1-t^2)-2t-2=0
3-3t^2 -2t-2=0
3t^2+2t-1=0
(3t-1)(t+1)=0
t=1/3,-1
Substituting back sin theta for t
sin theta=1/3 and sin theta=-1
Taking inverse on both sides
sin^-1(sin theta)=sin^-1(1/3) and sin^-1(sin theta)=sin^-1(-1)
theta= 19.5 degree, 160.5 degree and theta=-90degree, 270 degree
theta have to be greater than 0
therefore theta=19.5degree,160.5degree,270 degree
c. 5 sin theta+3=0
sin theta=-3/5
taking inverse on both sides
sin^-1(sin theta)=sin^-1(-3/5)
theta=sin^-1(-3/5)
theta=-36.87 degree,216.87degree, 323.13degree
And for theta lies between 0 and 360
theta=216.87 degree and 323.13 degree

