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Trig equations with radians.....

Solve 2sin (2x-Pi/3)=1 (0<X<2PI)

I simplified the equation to sin(2x-Pi/3)=1/2 (0.5)

That was a positive answer, so, using the CAST diagram, I identified that the solution laid in the 1st and 2nd quadrants (A and S).

1st Quadrant: 2x-Pi/3= Pi/6; (30 degrees) and 330 degrees (13 Pi over 6)

2nd Quadrant 2x+Pi/3=(210 degrees) or 5 Pi over 6;

That is as far as I got, before I had to draw a line under it, and declare myself officially lost. (NB I know the answers/solutions I have written thus far pertain to 2x; and would need to be halved to finalise the solution).

I had a look at the answers and I cannot make heads or tail of them.

For example

-Pi / 3 < 2x-Pi / 3 < 11 Pi over 3

11 Pi over 3 gives us 660 degrees as a result. I thought that this maybe due to the presence of the 2x-Pi/3; i.e. 2 x 360 (720)-60= 660 degrees. Is this correct. or just a fluke?

The answers also claim that the associated acute angle is the inverse of sin (1/2) to give us Pi over 6.

Why am I calculating the associated acute angle, and where did it come from?

I'm sorry if I somewhat garbled this message; I have tried to provide as much working/information as I could.
not sure why your answers are given as inequalities ?

:hmmmm2:

your working seems OK...

if you have 2x - 60º = 30º then x = 45º
Reply 2
Avatar for User947387
User947387
OP
17
Sorry the inequality was intended as "less than, or equal to 2 Pi"

i.e. to denote the range of the equation, (same way you get the likes of (0 less than or equal to x which is less than or equal to 360)
Reply 3
Avatar for ztibor
ztibor
10
Original post by apronedsamurai
Solve 2sin (2x-Pi/3)=1 (0<X<2PI)

I simplified the equation to sin(2x-Pi/3)=1/2 (0.5)

That was a positive answer, so, using the CAST diagram, I identified that the solution laid in the 1st and 2nd quadrants (A and S).

1st Quadrant: 2x-Pi/3= Pi/6; (30 degrees) and 330 degrees (13 Pi over 6)


for 13 Pi over 6 the sine is negative (that is in the 4th quadrant)
For the part solution i would use the 2 PI period ,
and I would limit the solutions in 0<X<2PI at the end.
2x−π3=π6+k⋅2π2x-\frac{\pi}{3}=\frac{\pi}{6}+k \cdot 2 \pi

2x=π2+k⋅2π 2x=\frac{\pi}{2}+k\cdot 2\pi

Dividing by 2 the period will be PI and more value of k (0,1) gives solution

x=π4+k⋅πx=\frac{\pi}{4}+k\cdot \pi


2nd Quadrant 2x+Pi/3=(210 degrees) or 5 Pi over 6;

The unknown angle in the 2nd quadrant is 2x-PI/3, too (this is in the original equation)
but here the sine of this angle will be 1/2 when this angle
is 5Pi/6 +k2Pi
Same as above
(edited 9 years ago)

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