find all the angles between 0° and 360° which satisfy the equation sin y = 3 cos (y - 30

sin y = 3 cos (y - 30)

siny = 3(cosy cos30 + siny sin30)
siny = 3(√3/2 cosy + 1/2 siny)
-1/2 siny = 3√3/2 cosy
tany = -3√3
tany < 0 in QII and QIV.

Assuming you are working in degrees ...

cos(y-30) = cosycos30 + sinysin30 = (√3/2)cosy + (1/2)siny
= (1/2)(√3cosy + siny)

sin y = 3 cos (y - 30)
siny = (3/2)(√3cosy + siny)
2siny = 3√3cosy + 3siny
siny = -3√3cosy
siny/cosy = -3√3
tany = -3√3
for your 0 ≤ y ≤ 360°
y = 100.9° or y = 280.9°

didn't realize oobleck already posted it,

must have happened while I was working it,
glad we have the same answer

Happens all the time mathhelper :)

To find all the angles between 0° and 360° that satisfy the equation sin(y) = 3cos(y - 30°), we can use the trigonometric identity:

sin(x - y) = sin(x)cos(y) - cos(x)sin(y).

Applying this identity to our equation, we get:

sin(y) = 3cos(y)cos(30°) + 3sin(y)sin(30°).

Now, we can rearrange the equation:

sin(y) - 3sin(y)sin(30°) = 3cos(y)cos(30°).

We can express sin(30°) and cos(30°) as known values:

sin(30°) = 1/2
cos(30°) = √3/2

Now, substituting these values into the equation, we have:

sin(y) - 3sin(y) * (1/2) = 3cos(y) * (√3/2).

Simplifying further:

sin(y) - (3/2)sin(y) = (3/2)√3 cos(y).

Multiplying both sides by 2 to remove the fractions:

2sin(y) - 3sin(y) = 3√3 cos(y).

Combining like terms:

-sin(y) = 3√3 cos(y) - 2sin(y).

Now, we can use the Pythagorean identity:

sin^2(x) + cos^2(x) = 1.

Rearranging this identity:

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

Substituting -sin(y) with √(1 - cos^2(y)):

√(1 - cos^2(y)) = 3√3 cos(y) - 2sin(y).

Next, we can square both sides of the equation to eliminate the square root:

1 - cos^2(y) = (3√3 cos(y) - 2sin(y))^2.

Simplifying further:

1 - cos^2(y) = (9 * 3 * cos^2(y)) + (4sin^2(y)) - (12√3 cos(y)sin(y)).

Expanding the equation:

1 - cos^2(y) = 27cos^2(y) + 4sin^2(y) - 12√3 cos(y)sin(y).

Rearranging terms:

0 = 26cos^2(y) + 4sin^2(y) - 12√3 cos(y)sin(y).

Now, we can substitute sin^2(y) as 1 - cos^2(y):

0 = 26cos^2(y) + 4(1 - cos^2(y)) - 12√3 cos(y)sin(y).

Simplifying further:

0 = 26cos^2(y) + 4 - 4cos^2(y) - 12√3 cos(y)sin(y).

Combining like terms:

0 = 22cos^2(y) + 4 - 12√3 cos(y)sin(y).

Now, we can factor out 2 from each term:

0 = 2(11cos^2(y) + 2 - 6√3 cos(y)sin(y)).

Since we want to find the angles between 0° and 360°, we need to solve this equation for y.

To simplify the process, let's divide each side by 2:

0 = 11cos^2(y) + 2 - 6√3 cos(y)sin(y).

Now, we're left with:

11cos^2(y) + 2 = 6√3 cos(y)sin(y).

Dividing both sides by cos^2(y):

11 + 2/cos^2(y) = 6√3 sin(y)/cos(y).

Using the trigonometric identity:

tan(y) = sin(y)/cos(y),

we can rewrite the equation as:

11 + 2/cos^2(y) = 6√3 tan(y).

Rearranging terms:

11cos^2(y) + 2 - 6√3 tan(y)cos^2(y) = 0.

Now, we have a quadratic equation in terms of cos(y). We can solve for cos(y) using the quadratic formula:

cos(y) = (-b ± √(b^2 - 4ac)) / (2a).

By comparing this to our quadratic equation, we can see that:

a = 11,
b = -6√3 tan(y),
c = 2.

Plugging these values into the quadratic formula, we can solve for cos(y) and then find the corresponding values of y.

Once we have the values of cos(y), we can substitute them back into the original equation sin(y) = 3cos(y - 30°), and solve for y.

Finally, we can check which values of y lie between 0° and 360° to find all the angles that satisfy the given equation.