Find all solutions to the equation. cos2x + 2 cos x + 1 = 0
write cos(2x) in terms of cos^2(x) using an identity then you would have a quadratic type equation
recall the double angle identity
cos^2x + 2 cos x + 1 = 0
cos(2x) doesn't equal cos^2(x)
use an identity for cos(2x)
no that's the original problem. the squared didn't show at first.
I took out a gcf of cosx and was left with cosx=0 and cosx=-3
do you know how to solve u^2+2u+1=0?
the gcf isn't cos(x) because cos(x) doesn't divide 1
no I don't , i got 3pi/2 ? but im not sure if its correct
do you know how to solve u^2+2u+1=0?
no
so you were never thought how to use the quadratic formula or how to factor?
taught*
yes i have been taught that. its just that we haven't done it in awhile so ts a little blurry
that is the way you are expected to solve this one
the coefficient of u^2 is 1 so all you need to answer is what two numbers multiply to be c and add up to be b in other words what two numbers multiply to be 1 and add up to be 2?
1 and 1
so u^2+2u+1 can be written as (u+1)(u+1) now back to solving u^2+2u+1=0 so we have (u+1)(u+1)=0 which means u has to be equal to?
-1
so if u=-1 and we solved u^2+2u+1=0 instead of cos^2(x)+2cos(x)+1=0 then that means u=cos(x) which means cos(x)=-1
can you solve this equation for x cos(x)=-1?
ohhh so it would be (pi)
i understand it now. thank you
pi is a solution between 0 and 2 pi there are many more solutions outside that domain of numbers
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