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
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-> 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
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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.
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