Question

Please solve the equation on the interval 0 (less-than or equal-to) theta < 2pi

sin(2theta) = -(sqrt3)/2

Thank you

Answer 1

If you add fish to the equation then you end up with cat x 7 / 3

Answer 2

5pi/6

Answer 3

well using tri ratios

\[2\theta = \sin^{-1} (\sqrt{3}/2) = \pi/3\]

sin is negative in 3rd quadrant ( pi + theta) and 4th quadrant ( 2pi - theta)

so the solutions are

\[\theta = 4\pi/3, 5\pi/3\]

Answer 4

2θ = π/3 --> θ = π/6

Answer 5

How did I get 2pi/3 and 5pi/3

Answer 6

oops should be pi/6.... forgot to have it

Answer 7

sin is negative... thats the key............. in the 3rd quadrant and 4th quadrant sin(x) is negative... so find theta

then 3rd quadrant is (pi + theta) 4th quadrant is (2pi - theta)

Answer 8

Oh no...i don't believe it, i typed the equation incorrectly....sqrt3/2 is not negative, it is positive

Answer 9

im so sorry

Answer 10

:P

Answer 11

That's why you have different answer from ours!

Answer 12

type into your calculator sin(7pi/6) and (11pi/6) is the answer

\[-\sqrt{3}/2\]

Answer 13

then its 1st and 2nd quadrants where sin is positive

so the angles are pi/6 and pi - pi/6 = 5pi/6

Answer 14

i got theta=pi/6 and 7pi/6

Answer 15

@Mike, if you're happy with the solver, click on the blue Best answer, will you :)

Answer 16

ummmm good luck

Answer 17

thanks