Question

Help!!

Answer 1

\[3\sin \theta = \sin \theta -1\]

Answer 2

sin(theta)= -1/2->theta=k*pi - (-1)^k*(pi/6) |k belongs to Z

Answer 3

@Pologirl19 get it?

Answer 4

When I do sin^-1 of (-1/2), I am getting - pi/6 as one of my answers, which when I look at the unit circle actually confuses me. does this mean it is really pi/6 as one of my answers?

Answer 5

this is a multiple choice and none of my answers have a negative sign

Answer 6

no pi/6 is not a answer

Answer 7

\[3\sin \theta-\sin \theta=-1,\sin \theta=\frac{ -1 }{ 2 }=-\sin \frac{ \pi }{ 6 }\]
\[\sin \theta=\sin \left( \pi+\frac{ \pi }{ 6 } \right),\sin \left( 2\pi-\frac{ \pi }{6 } \right) \]
\[\theta=\frac{ 7\pi }{ 6 },\frac{ 11\pi }{6 }\]
\[adding 2n \pi \to above solution we get infinite solutions ,where nis an integer.\]

Answer 8

k=0 ->-pi/6

Answer 9

k=1->7*pi/6

Answer 10

K=2 -> 11*pi/6

Answer 11

and so on

Answer 12

okay, now looking at both of your answers, that makes sense, and I had originally answered it as 7pi/6 and 11pi/6, but I was missing a piece of the puzzle of how i came to that answer. thank you

Answer 13

Can you help me with another one?
\[\tan^2 \theta = -3/2 \sec \theta \]

Answer 14

@RaphaelFilgueiras and @surjithayer

Answer 15

try to use this sec²x=1+tg²x, you will have a second degree equation with secx as variable

Answer 16

I need to use pythagorean identity of 1+tan^2 =sec^2?

Answer 17

yes,
\[\tan^2 \theta = -3/2 \sec \theta= tan^2\theta*sec\theta=-3=tan^2\theta(1+tan^2\theta)=-3\]

Answer 18

expand and move -3 to the LHS and solve as polynomial with variable is tan (theta)

Answer 19

I got 2pi/3 and 4pi/3. Is that right?

Answer 20

@RaphaelFilgueiras can you check for her?
@Pologirl19 I am sorry friend, I didn't work for the answer yet. just way to do