+956 Hello,
I've been struggling a bit with where to start off from here. If someone could guide me that would be great!
Solve \(tan^2sinx -sinx/3\) = 0 on the interval 0 ≤ x ≤ 2π
The way I approached this was try to get rid of the fractions so I multiplied each side by 3. I then found my GCF of sinx and factored it through. It looked like this: sinx(3tan^2x-1)
Am I correct? Any ideas thoughts?
+128407 tan^2 (x) * sin (x) - sin (x) / 3 = 0 get a common denominator
[ 3tan^2 (x) * sin (x) - sinx] / 3 = 0 mutiply both sides by 3
3tan^2 (x) * sin (x) - sin (x) = 0 factor out sin(x)
sin (x) [ 3tan^2 ( x) - 1] = 0
We have two equations here.....either.......
sin (x) = 0 and this happens at 0 and pi
Or
3tan^2 (x) - 1 = 0 add 1 to both sides
3tan^2 (x) = 1 divide both sides by 3
tan^(2) x = 1/3 take both roots
tan (x) = 1/√3 and this happens at pi/6 and 7pi/6
And
tan (x) = - 1/√3 and this happens at 5pi/6 and 11pi/6
So.....the solutions are 0, pi/6, 5pi/6, pi, 7pi/6 and 11pi/6
+128407 tan^2 (x) * sin (x) - sin (x) / 3 = 0 get a common denominator
[ 3tan^2 (x) * sin (x) - sinx] / 3 = 0 mutiply both sides by 3
3tan^2 (x) * sin (x) - sin (x) = 0 factor out sin(x)
sin (x) [ 3tan^2 ( x) - 1] = 0
We have two equations here.....either.......
sin (x) = 0 and this happens at 0 and pi
Or
3tan^2 (x) - 1 = 0 add 1 to both sides
3tan^2 (x) = 1 divide both sides by 3
tan^(2) x = 1/3 take both roots
tan (x) = 1/√3 and this happens at pi/6 and 7pi/6
And
tan (x) = - 1/√3 and this happens at 5pi/6 and 11pi/6
So.....the solutions are 0, pi/6, 5pi/6, pi, 7pi/6 and 11pi/6
