How do I solve this precalc problem?
Mathematics

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sin2a  cosa = 0. Find all solutions of the equation in the interval [0, 2 π)
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Ok, first we use this trigonometric identity (double angle formula for the sine function) which is:
sin(2a) = 2sin(a)cos(a)
So we enter it into the original expression resulting:
sin(2a) cos(a) = 0 > 2sin(a)cos(a)  cos(a) = 0
So here we can see that we have a common factor which is the cos(a) since we have it twice in our expression:
2sin(a)cos(a)  cos(a) = 0
Then we factor cos(a) out from the whole expression like this:
cos(a)(2sin(a)  1) = 0
After this, we equal each factor to zero and solve for a. So we would have the following:
cos(a) = 0 and 2sin(a)  1 = 0
So we solve the first equation for a by taking the inverse function from both sides (which is arccos).
cos(a) = 0 > arccos(cos(a)) = arccos(0) > a = arccos(0)
In order to get the arccos(0) we need to find the angles or radians at which the cosine function is equal to zero.
So let's remember the cosine function is zero at the the odd multiples of Pi/2 which are: Pi/2, 3Pi/2, 5Pi/2, 7Pi/2...
Since it says that we just consider the solutions between the interval [0 , 2Pi) then we just consider
a = Pi/2 and a = 3Pi/2
Now we solve the other equation which is: 2sin(a)  1 = 0. So first, we add 1 from both sides like this:
2sin(a)  1 = 0 > 2sin(a)  1 + 1 = 0 + 1 > 2sin(a) = 1
Then we divide by 2 from both sides:
2sin(a) = 1 > 2sin(a)/2 = 1/2 > sin(a) = 1/2
Finally, we take the inverse of the sine function from both sides (the arcsin):
sin(a) = 1/2 > arcsin(sin(a)) = arcsin(1/2) > a = arcsin(1/2)
So in order to find the arcsin(1/2) we need to find the angles or radians at which the sine function is equal to 1/2.
Then we have: a = Pi/6. In order to get the other solution we just subtract Pi from the first solution. So we have:
a = Pi  Pi/6 = Pi(1  1/6) = Pi(6/6  1/6) = Pi(5/6) = 5Pi/6.
So finally, the solutions are: a = Pi/6 , a = 5Pi/6 , a = Pi/2 , a = 3Pi/2
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