Question 627380: Use an algebraic method to find the exact solution(s) of 2 cos^2x + sin x =1 where x is any real number. (Please show work)
Answer by hammy(9)   (Show Source):
You can put this solution on YOUR website! In this particular case, you should try and set up the equation in quadratic form so that you can solve for your variable by simply factoring and applying the zero product property. First, we want to rewrite the equation so that we only use sin or cos. So,
cos^2x=1-sin^2x by the Pythagorean identity
Substitute into the original equation
2(1-sin^2x)+sinx=1
2-2sin^2x+sinx-1=0 Notice that we are setting it to zero in order to factor.
We also have one sin squared and one not and a constant.
The main goal is to make it look like this
ax^2+bx+c
so,
-2sin^2x+sinx+1=0
To make things easier, make the substitution sinx=y
-2y^2+y+1=0
-(2y^2-y-1)=0
-[(2y^2-2y)+(y-1)]=0
-[2y(y-1)+(y-1)]=0
-(2y+1)(y-1)=0
By the zero product property
y=-1/2 and y=1
sinx=-1/2 and sinx=1
x=7pie/6+2pie*n,11pie/6+2pie*n,pie/2+2pie*n
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