find all real numbers in this interval (o, 2 pi) that satisfy this equation. round to the nearest tenth. cos x(square root symbol) 3 = 3(square...

If the equation is cos x*sqrt 3 = 3*sqrt (3*sin x), we'll
raise to square both sides to remove the square root from both
sides:

3(cos x)^2 = 9*3*sin
x

We'll divide by 3 both
sides:

(cos x)^2 = 9*sin
x

We'll use the Pythagorean identity to express cos x with
respect to sin x:

(cos x)^2 = 1 - (sin
x)^2

1 - (sin x)^2 = 9*sin
x

We'll subtract 9sin x both
sides:

- (sin x)^2 - 9*sin x + 1 =
0

(sin x)^2 + 9*sin x - 1 =
0

We'll replace sin x by
t:

t^2 + 9t - 1 = 0

We'll
apply quadratic formula:

t1 = [-9 + sqrt(9^2 -
4*1*(-1))]/2*1

t1 =
(-9+sqrt85)/2

t1 = 0.1097

t2 =
(-9-sqrt85)/2

t2 = -9.1097

sin
x = t1 => sin x = 0.1097 => x = arcsin 0.1097 => x = 6.2980 degrees
(1 st quadrant) or x = 180 - 6.2980 = 173.702 degrees (2 nd
quadrant)

sin x = -9.1097 impossible since the values of
the function sine cannot be smaller than
-1.

Therefore, the solutions of the equation
are expressed in degrees and they are: {6.2980 ;
173.702}.