Solve this equation fo rx in the interval 0<=x<=360

3sinxtanx=8

I would do it this way:
sinxtanx = 8/3
sinx(sinx/cosx)=8/3
sin^2x/cosx = 8/3
(1-cos^2x)/cosx=8/3
cross-multiply
3 - 3cos^2x = 8cosx
3cos^2x + 8cosx - 3 = 0
(3cosx-1)(cosx+3)=0
cosx=1/3 or cosx = -3
the last one is not possible
so x = arccos(1/3)=70.5º
or 289.5º (in the fourth quadrant)

To solve the equation 3sinxtanx = 8 for x in the interval 0 <= x <= 360, follow these steps:

1. Divide both sides of the equation by 3 to isolate sinx * tanx.
sinx * tanx = 8/3

2. Rewrite sinx * tanx as sinx * (sinx / cosx) using the identity tanx = sinx / cosx.
sinx * (sinx / cosx) = 8/3

3. Simplify the equation further by multiplying sinx:
(sin^2x) / cosx = 8/3

4. Multiply both sides of the equation by cosx to eliminate the denominator:
(sin^2x) = (8/3) * cosx

5. Rewrite sin^2x as (1 - cos^2x) using the Pythagorean identity sin^2x + cos^2x = 1:
(1 - cos^2x) = (8/3) * cosx

6. Distribute (8/3) to both terms on the right side:
1 - cos^2x = (8/3) * cosx

7. Move all terms to one side of the equation:
1 - (8/3) * cosx - cos^2x = 0

8. Rearrange the quadratic equation in standard form:
cos^2x + (8/3) * cosx - 1 = 0

9. Factor the quadratic equation:
(3cosx - 1)(cosx + 3) = 0

10. Set each factor equal to zero and solve for cosx:
3cosx - 1 = 0 or cosx + 3 = 0

11. Solve the first equation, 3cosx - 1 = 0, for cosx:
3cosx = 1
cosx = 1/3

12. Solve the second equation, cosx + 3 = 0, for cosx:
cosx = -3

13. Since cosx cannot be -3 in the interval 0 <= x <= 360, we discard that solution.

14. To find the values of x, take the inverse cosine (arccos) of cosx:
x = arccos(1/3) = 70.5º

15. Since cosine is positive in both the first and fourth quadrant, the second solution is:
x = 360º - 70.5º = 289.5º

The two solutions for x in the interval 0 <= x <= 360 are 70.5º and 289.5º.