Question

Find the values of thetha between 0 and 2 pi that satisfy sin thetha = - square root 3 / 2

Answer 1

got your Unit Circle handy?

Answer 2

sin is negative in third and fourth quadrant.

Answer 3

\(\bf sin(\theta)=-\sqrt{\cfrac{3}{2}}\qquad \qquad 0<\theta<2\pi
\\ \quad \\
sin^{-1}[sin(\theta)]=sin^{-1}\left( -\sqrt{\cfrac{3}{2}} \right)\implies \theta=sin^{-1}\left( -\sqrt{\cfrac{3}{2}}\right)\)

Answer 4

yes i do

Answer 5

so check your unit circle for "those angles" within that constrained range

Answer 6

and keep in mind what surjithayer pointed out
you want a negative sine, and that means, the 3rd and 4th quadrants

Answer 7

Thank you so much! Both of you! :)

Answer 8

yw

Answer 9

\[\sin \theta=-\frac{ \sqrt{3} }{ 2 }=-\sin \frac{ \pi }{ 3 }=\sin \left( \pi+\frac{ \pi }{ 3 } \right),\sin \left( 2\pi-\frac{ \pi }{ 3} \right)\]

Answer 10

if it is\[\sin \theta=-\sqrt{\frac{ 3 }{ 2 }}\]
then the calculations of jdoe0001 are correct.

Answer 11

it is \[\sin \theta = -\frac{ \sqrt{3}}{ 2 }\] @surjithayer

Answer 12

\[\sin \theta=\sin \frac{ 4 \pi }{ 3 },\sin \left( \frac{ 5 \pi }{ 3 } \right),\theta=?\]

Answer 13

Is it .091?

Answer 14

no
\[\theta=\frac{ 4 \pi }{ 3 },\frac{ 5 \pi }{ 3 }\]

Answer 15

you have to find the angles