SOLUTION: 2sin^2x+3cosx=3
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Question 252486: 2sin^2x+3cosx=3
Found 2 solutions by drk, jim_thompson5910:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Here is the original problem:
(i) 2sin^2x+3cosx=3
We have an identity for sin^2(x).
Sin^2(x) + Cos^2(x) = 1.
Solving for Sin^2(x), we get
sin^2(x) = 1 - cos^2(x)
Now, by substitution into (i), we get
2(1 - cos^2(x)) + 3cos(x) - 3 = 0
2cos^2(x) - 3cos(x) + 1 = 0.
Factoring, we get
(2cos(x) - 1)(cos(x) - 1) = 0.
Solve each parenthesis for x.
(2cos(x) - 1) = 0
cos(x) = 1/2.
x = 60 degrees or 300 degrees ; radians = pi/3, 5pi/3
(cos(x) - 1) = 0
cos(x) = 1
x = 0 degrees ; 0 radians.
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So, your answers are: 0, 60 degrees, 300 degrees.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Hint: Recall that +cos^2(x)=1)
.
So this means that =1-cos^2(x))
.
So plug in to get \right)+3cos(x)=3)
.
Now let )
and plug that in to get: +3z=3)
.
At this point, we have a quadratic which can be solved by the quadratic formula. You'll get solutions in terms of 'z', so you'll need to remember to use to find the solutions in terms of 'x'. Let me know if you still need help.
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