If cos 0 = 0.3846 where 3pi/2 is less than 0 is less than 2pi, what is the approximate value of sin 0?

-0.9231

Step-by-step explanation:

Given cos theta = 0.3846 within the range 3π/2<theta<2π

Costheta = 0.3846

Theta = arccos(0.3846)

Theta = 67.38°

Since 67.38° didn't fall within the range of theta, we will locate our theta using the quadrant.

Since cos is positive in the 4th quadrant,

Theta = 360-67.38°

Theta = 292.62°

Sin(theta) = sin(292.62°)

= -0.9231

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)