Question

Another Trigonometry question Help!

Solve the equation in the interval [0°,360°).
4 sin^2θ=3


Please show your steps thank you :)

Answer 1

Ok so first step is to divide 4 on both sides:
\[\sin^2(\theta)=\frac{3}{4}\]
Second step is to get rid of that square on the left side so take square root of both sides:
\[\sin(\theta)=\pm \sqrt{\frac{3}{4}}\]

Answer 2

4 sin^2θ=3

sin(t)^2=3/4 divide by 4

sin(t)=sqrt (3/4) take sqrt of both sides

t=arcsin(sqrt (3/4)) take arc sin

Answer 3

\[\sin(\theta)=\frac{\sqrt{3}}{2} \text{ or } \sin(\theta)=-\frac{\sqrt{3}}{2}\]
Use unit circle :)

Answer 4

Hmm..i am actually confused on what the questions is asking for....i understand your guys work, i am just lost on what the question is asking

Answer 5

It wants you to find all values of theta that could satisfy the equation in the interval given

Answer 6

ohhh, cause so when i use my unit circle i get... sin 30°, sin 150°, sin 210° and sin 330° ....is this right?

Answer 7

\[\sin(30) \neq \frac{\sqrt{3}}{2}\]
\[\sin(150) \neq \frac{\sqrt{3}}{2}\]
\[\sin(210) \neq \frac{-\sqrt{3}}{2}\]
\[\sin(330) \neq \frac{-\sqrt{3}}{2}\]

Answer 8

\[\sin(60)=?\]
\[\sin(120)=?\]
\[\sin(240)=?\]
\[\sin(300)=?\]

Answer 9

now I understand...okay i was looking at the ones in the x coordinate instead of the y coordinate...got it now

Answer 10

what about sin 45 that would work also right?

Answer 11

no
\[\sin(45)=\frac{\sqrt{2}}{2} \neq \frac{\sqrt{3}}{2} \text{ or } \frac{-\sqrt{3}}{2}\]

Answer 12

you are looking for where you have that the sine value is
either \[\frac{\sqrt{3}}{2} \text{ or } \frac{-\sqrt{3}}{2}\]

Answer 13

I see....now I understand, thanks again!!