Question

how to solve for x in this trig equation
$3sin(2x)-2sinx=0$

thanks

Answer 1

Explanation:

rearrange euating to zero

⇒2sin2x+sinx−1=0

we now have a quadratic in sin

⇒(2sinx−1)(sinx+1)=0

⇒2sinx−1=0 or sinx+1=0

⇒sinx=12 or sinx=−1

Answer 2

Step-by-step explanation:

The key is to make a substitution. Let u = sin x. Then you have:

3u^2 - 2u = 0

u(3u - 2) = 0

u = 0 or 3u - 2 = 0

u = 0 or u = 2/3

Now, unsubstitute with u = sin x , and you get:

sin x = 0 or sin x = 2/3

For the first equation, you can get exact answers. Sin x = 0 when x = 0, 180, 360, .... = 180k for integer k.

For the second equation, you have to take the arcsine of 2/3 by a calculator. Then, you know that that will be an acute angle. Take that angle on a unit circle and reflect it across the y-axis. The sine of that angle will STILL be 2/3. So, instead of theta degrees above the +x axis, it's theta degrees above the -x axis, or in other words it's a 180 - theta degree revolution.

So, if arcsine 2/3 were to be 55 degrees (it's probably not), then your two angles would be 55 degrees and 180 - 55 = 125 degrees.