SOLUTION: how do you solve for x algebraically for : 2sin^2x+5sinx=3

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Question 784570: how do you solve for x algebraically for : 2sin^2x+5sinx=3
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
2%28sin%28x%29%29%5E2%2B5sin%28x%29=3
If you make a change of variable, it gets easier to write and easier to see the solution.
Substititing y=sin%28x%29, the equation turns into
2y%5E2%2B5y=3
You probably know what to do with that equation.
I would do this:
2y%5E2%2B5y=3 --> 2y%5E2%2B5y-3=0 --> %282y-1%29%28y%2B3%29=0
From that last, factored equation, we get two solutions in terms of y:
2y-1=0 --> 2y=1 --> y=1%2F2 and
y%2B3=0 --> y=-3
In terms of x, y=sin%28x%29=-3 makes no sense.
On the other hand, from the other y value we get many values for x
y=sin%28x%29=1%2F2
To begin with, if we just stick to the first turn, with 0%5Eo%3C=x%3C360%5Eo, we have 2 solutions: highlight%2830%5Eo%29 and highlight%28150%5Eo%29, because
sin%2830%5Eo%29=1%2F2 and sin%28150%5Eo%29=1%2F2
If using radians rather than degrees, for 0%3C=x%3C2pi, the soluttions are highlight%28pi%2F6%29 and highlight%285pi%2F6%29, because
sin%28pi%2F6%29=1%2F2 and sin%285pi%2F6%29=1%2F2
If you are not restricted to the first turn, there are infinite solutions, because co-terminal angles have the same trigonometric function values, so that all the solutions could be written as
%284k%2B1%2990%5Eo+%2B-+60%5Eo or %284k%2B1%29pi%2F2+%2B-+pi%2F3 for any integer k.
(Making k=0 gives you the solutions in the first turn).