Encountered a little problems with this question. Please take a look.(=
find all the angles between 0 deg and 360 deg which satisfy the equation
2sin y = tan y
cos y = 2
y lies in first or fourth quadrant.
cos alfa = 1/2
alfa = 60 deg
y - 60 deg, 300 deg
I am missing on 180 deg(stated in the answers). how can you find that?
THANKS!
2sin y = tan y
2sin y - sin y / cos y = 0
sin y (2cos y - 1) = 0
either siny = 0 or cos y = 1/2
From sin y = 0, you get your 180 degrees.
Remember for such equations, never divide the sin y away. You will lose solutions, as shown clearly by your example.
Another example would be 2x^2 - x = 0
You don't divide by x to get 2x -1 = 0. Instead, you factorise to x (2x - 1) = 0. The same thing is to be used on your question here.
Regards,
Eagle
Owner of Strategic Tuition