Find all the angles between 0 and 360 degree which satisfy the equation. (i) sin x+sin60 degree=0. (ii) tan 2y= -0.5.

Answer 1

have

sin(x)=sin(-60)

x=-60+k*360

and

x=180-(-60)+k*360

use k=1 in first equation and k=0 in the second equation.

x=240,300

note sin(240)=sin(300)=sin(-60)=-sin(60)

have

2y=tan^-1(-1/2)+180*k,,,,k=0,+/-1,...

y=[tan^-1(-1/2)]/2]+90*k=-13.18+90*k approx.

k=1 y=76.82

k=2 y=166.82

k=3 y=256.82

k=4 y=346.82

k=1,,check tan(153.64)=-.496

Answer 2

sinx + sin60 = 0

sin60 = sqr3/2

sinx + sqr(3/2) = 0

sinx = 0 - sqr3/2 = -sqr3/2

x is in quadrant III or IV where sines are negative

reference angle equal 60 degrees

x = sin^-1(-sqr3/2) = about 180+ 60 or 360-60

x = 240 or 300

tan2y = -.5 = -1/2

angle 2y is in quadrant II or IV where tangents are negative

reference angle = tan^-1(-1/2) = about 26.565

2y = tan^-1(-1/2) = about 180- 26.565 or 360-26.565, with each also added to 360

2y = about 153.435, 333.435, 513.435, 693.435

y = about 76.7, 166.7, 256.7, or 346.7