Question

Solve each of the following trigonometric equations for 0<=x<=360
a) cosec x= 9 sin x
@ganeshie8

Answer 1

just to double check my answer

Answer 2

\[cosec~x=9~\sin~x\]\[\frac{ 1 }{ \sin~x }=9~\sin~x\]\[\sin^2~x=\frac{ 1 }{ 9 }\]\[\sin~x=\frac{ 1 }{ 3 }\]\[x=19.07,160.53\]

Answer 3

*19.47

Answer 4

i have few mints i have to go somewhere but i don't how u did u get 160
but when we take square root \[\rm \sqrt{\sin^2x}=\sqrt{\frac{1}{9}} \rightarrow sinx= \pm \frac{1}{3}\]
so take invs of sin sin^{-1}(-1/3) = -19.47
and use the fact circle add up to360
to get the positive angle add 360 with -19.47

Answer 5

oh i totally forget about the = and -...
Thank you @Nnesha

Answer 6

*+

Answer 7

so the answer should be \[\sin~x=\frac{ 1 }{ 3 }~and~\sin~x=-\frac{ 1 }{ 3 }\]\[x=19.47,160.53~and~x=(180+19.47),(360-19.47)\]\[x=19.47,160.53,199.47,340.53\]am i correct? @Nnesha