# How do you solve for x when 2sin(2x-60°)=1?

• Apr 7th 2009, 08:43 PM
Jd09
How do you solve for x when 2sin(2x-60°)=1?
How do you solve for x for this equation: 2sin(2x-60°)=1?

where -2π ≤ x ≤ 2π

[π=pi]

I am unsure of where to start for this one (Shake)
• Apr 7th 2009, 08:48 PM
Jhevon
Quote:

Originally Posted by Jd09
How do you solve for x for this question: 2sin(2x-60°)=1?

where -2π ≤ x ≤ 2π

I am unsure of where to start for this one (Shake)

$2 \sin (2x - 60) = 1 \implies \sin (2x - 60) = \frac 12$ $\implies 2x - 60 = \sin^{-1} \left( \frac 12 \right) \implies x = \frac {\sin^{-1}(1/2) + 60}2$

so now what? what are all the solutions? which have x in the given range?
• Apr 7th 2009, 09:21 PM
Jd09
Is this correct?

(sin-1)sin(2x-60)=1/2(sin-1)

(2x-60)=30
x=45

??
• Apr 7th 2009, 09:38 PM
Jhevon
Quote:

Originally Posted by Jd09
Is this correct?

(sin-1)sin(2x-60)=1/2(sin-1)

(2x-60)=30
x=45

??

that is correct...is that the only solution? you want $-360^o \le x \le 360^o$ remember?
• Apr 7th 2009, 09:48 PM
Jd09
135...

and would -135 work as well?
• Apr 7th 2009, 09:53 PM
Jd09
How do you solve for x when 2sin(2x-60°)=1?