# why equation, wrong

• Jan 15th 2011, 12:25 PM
paulaa
why equation, wrong
4 sin^2(x) + 8 cos(x) = 7

1 = cosē(x)+sinēx
sinē(x) = 1-cosē(x)

we use this and replace sinē(x) with 1-cosē(x)

4(1-cosē(x))+8cos(x) = 4-4cosē(x)+8cos(x)

-4cosē(x)+8cos(x)+4 = 7

replace
[cos(x)=t] and multiplicate with -1

4tē-8t-4 = -7 <--> t1= 1/2 t2=3/2

t = cos(x),

cos(x1)=1/2
cos(x2)=3/2

x1=pi/3 + n *2pi or x1=pi - pi/3 + n *2pi but the second one is wrong ? why ?
Cos(x2) > 1 = dont work
• Jan 15th 2011, 12:31 PM
e^(i*pi)
You're working is fine - it's a domain and range issue - you can discard t2

The maxiumum value of $\cos(x)$ is 1. Since $\dfrac{3}{2} > 1$ then you cannot take the inverse cos.

It's not in the range of $f(x) = \cos(x)$ and because cos(x) and arccos(x) are inverses of each other then the range of $f(x)$ becomes the domain of $f^{-1}(x)$ which means it's not in the domain of $f^{-1}(x)$
• Jan 15th 2011, 01:07 PM
paulaa
yes i know that i meant that the wrong is in x1=pi - pi/3 + n *2pi
• Jan 15th 2011, 01:10 PM
paulaa
i think maybe it will be x1=pi/3 + n *2pi or x1=pi - pi/3 - n *2pi

right now ?
• Jan 16th 2011, 05:44 PM
SammyS
Quote:

Originally Posted by paulaa
4 sin^2(x) + 8 cos(x) = 7
...
cos(x1)=1/2

x1=pi/3 + n *2pi or x1=pi - pi/3 + n *2pi but the second one is wrong ? why ?

cos(x) is an even function. If cos(x1) = 1/2, then also cos(-x1) = 1/2

Therefore the second one should be: -x1 = pi/3 → x1 = -p1/3. Of course, you can add n*2pi
So the final form for the second solution is: x1 = pi/3 + n 2pi.