SOLUTION: Solve the following equation over the domain 0 < x < 2pie
2cos^2 x = 1 + sin x
I am getting the solution wrong.
I thought it would be this:
2(1 - sin^2 x) = 1 + sin
Algebra.Com
Question 596622: Solve the following equation over the domain 0 < x < 2pie
2cos^2 x = 1 + sin x
I am getting the solution wrong.
I thought it would be this:
2(1 - sin^2 x) = 1 + sin x
2 - 2sin^2 x = 1 + sin x
0 = 2sin^2 x + sin x - 1
0 =(2sin x - 1)(sin x + 1)
using the null factor law
either 2sin x - 1 = 0 OR sin x + 1 = 0
sin x =1/2 sin x = -1
Where sin x = 1/2 is equivalent to pie(3.14)/6 (30 degrees)
Sine is positive in the 1st and 2nd quadrants so it will be pie(3.14)/6 and 5pie(3.14)/6 and sin x = -1 at 3pie(3.14)/2
So i thought the solutions are pie/6, 5pie/6 and 3pie/2?
Any help with this would be great.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
Why do you think your solution is wrong?
The only thing missing, to me, is that you should first write the general solution:
From these we find the specific solutions which are in the specified interval. And you are right on the money with the specific solutions:
or
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