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Hi! I hope you don't mind that I have another question! Your answer to my last question was really helpful! How would I solve this trigonometric equation for 0≤x≤2pi?
(2(sin^2)x)+(sinx)-1=0
Thanks!!
Hello again Hannah,
To make the algebra *look* easier, substitute u=sin(x).
So, the "new" problem is 2u^2+u-1=0...now factor!
(2u-1)(u+1)=0 ==> 2u-1=0 OR u+1=0 ==> u=1/2 OR u=-1
Now put sin(x) back in for u, to get: sin(x)=1/2 OR sin(x)=-1
sin(x)=1/2 gives x=pi/6 or 5pi/6
sin(x)=-1 gives x=3pi/2
I hope this helps you again!
BTW: What class is this for? a HS or college class?
TTYL, Abe
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| Rating(1-10) | Knowledgeability = 10 | Clarity of Response = 10 | Politeness = 10 |
| Comment | Thanks!! It did help again! It's for an AP Calculus AB high school class. We have to do the prerequisite chapters in our book over the summer so we can have more time in class to prepare for the AP test. Thanks again so much for your help!! | ||
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Calculus
Answers by Expert:
Abe Mantell
Expertise
Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Experience
Over 15 years teaching at the college level.
Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.
Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook
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