Find all the angles between 0° and360° which satisfy: 2sin^2 y - 3cos y = 0?
To find the angles between 0° and 360° that satisfy the equation 2sin^2(y) - 3cos(y) = 0, we can use trigonometric identities to simplify the equation.
Let's start by using the identity sin^2(y) + cos^2(y) = 1. Rearranging this equation, we have sin^2(y) = 1 - cos^2(y).
Substituting this into the original equation, we get:
2(1 - cos^2(y)) - 3cos(y) = 0.
Expanding the equation, we have:
2 - 2cos^2(y) - 3cos(y) = 0.
Rearranging the terms, we can rewrite it as:
2cos^2(y) + 3cos(y) - 2 = 0.
Now we can solve this quadratic equation. We can factor it as:
(2cos(y) - 1)(cos(y) + 2) = 0.
Setting each factor equal to zero, we have two separate equations:
2cos(y) - 1 = 0,
cos(y) + 2 = 0.
Solving the first equation, we find:
2cos(y) = 1,
cos(y) = 1/2.
Using the unit circle or a calculator, we know that cos(y) = 1/2 has solutions at y = 60° and y = 300°.
For the second equation, we have:
cos(y) = -2.
However, there are no real solutions for cos(y) = -2 between 0° and 360° since the cosine function only takes values between -1 and 1.
Therefore, the angles between 0° and 360° that satisfy the equation 2sin^2(y) - 3cos(y) = 0 are y = 60° and y = 300°.
To find all the angles between 0° and 360° that satisfy the equation 2sin^2(y) - 3cos(y) = 0, we can use the trigonometric identity sin^2(y) + cos^2(y) = 1.
Let's rearrange the equation by substituting sin^2(y) = 1 - cos^2(y):
2(1 - cos^2(y)) - 3cos(y) = 0.
Simplifying the equation, we get:
2 - 2cos^2(y) - 3cos(y) = 0.
Rearranging the terms, we have:
2cos^2(y) + 3cos(y) - 2 = 0.
Now we can solve this quadratic equation for cos(y) using factoring or the quadratic formula.
By factoring, we find:
(2cos(y) - 1)(cos(y) + 2) = 0.
Setting each factor equal to zero, we get:
2cos(y) - 1 = 0 or cos(y) + 2 = 0.
Solving each equation separately:
2cos(y) - 1 = 0
2cos(y) = 1
cos(y) = 1/2
y = arccos(1/2) ≈ 60° or 300°.
cos(y) + 2 = 0
cos(y) = -2 (reject due to being outside the range of [-1, 1]).
Therefore, the angles y that satisfy the equation 2sin^2(y) - 3cos(y) = 0 between 0° and 360° are approximately 60° and 300°.
substituting ... 2[1 - cos^2(y)] - 3 cos(y) = 0 ... -2 cos^2(y) - 3 cos(y) + 2 = 0
solve the quadratic for cos(y) and fill in the angles