the graph of y = sin x and the line y = (-1)/2 over the interval [0 degree, 360 degree]. Where do the two graphs intersect? Give exact answers in degrees
+26367 the graph of y = sin x and the line y = (-1)/2 over the interval [0 degree, 360 degree]. Where do the two graphs intersect? Give exact answers in degrees
$$\small{\text{$ \begin{array}{rcl|rcl}\sin{(x_1)} &=& -\frac12 \qquad & \qquad \sin{(x_1)} &=& \sin{ (180\ensurement{^{\circ}} -x_2) } =-\frac12 \\&&&\\ x_1 &=& \arcsin{( -\frac12 )} \qquad & \qquad \sin{ (180\ensurement{^{\circ}} - x_2) } &=&-\frac12 \\&&&\\x_1 &=& -30\ensurement{^{\circ}} + 360\ensurement{^{\circ}} =330\ensurement{^{\circ}} \qquad & \qquad 180\ensurement{^{\circ}} - x_2 &=& \arcsin(-\frac12) \\&&&\\&& &\qquad x_2 &=& 180\ensurement{^{\circ}}-\arcsin{( -\frac12 )} \\&&&\\&&\qquad & x_2 &=& 180\ensurement{^{\circ}} +30\ensurement{^{\circ}} \\&&&\\&&\qquad & x_2 &=& 210\ensurement{^{\circ}}\\\end{array}$}}$$
+26367 the graph of y = sin x and the line y = (-1)/2 over the interval [0 degree, 360 degree]. Where do the two graphs intersect? Give exact answers in degrees
$$\small{\text{$ \begin{array}{rcl|rcl}\sin{(x_1)} &=& -\frac12 \qquad & \qquad \sin{(x_1)} &=& \sin{ (180\ensurement{^{\circ}} -x_2) } =-\frac12 \\&&&\\ x_1 &=& \arcsin{( -\frac12 )} \qquad & \qquad \sin{ (180\ensurement{^{\circ}} - x_2) } &=&-\frac12 \\&&&\\x_1 &=& -30\ensurement{^{\circ}} + 360\ensurement{^{\circ}} =330\ensurement{^{\circ}} \qquad & \qquad 180\ensurement{^{\circ}} - x_2 &=& \arcsin(-\frac12) \\&&&\\&& &\qquad x_2 &=& 180\ensurement{^{\circ}}-\arcsin{( -\frac12 )} \\&&&\\&&\qquad & x_2 &=& 180\ensurement{^{\circ}} +30\ensurement{^{\circ}} \\&&&\\&&\qquad & x_2 &=& 210\ensurement{^{\circ}}\\\end{array}$}}$$
