# Math Help - why equation, wrong

1. ## why equation, wrong

4 sin^2(x) + 8 cos(x) = 7

1 = cos²(x)+sin²x
sin²(x) = 1-cos²(x)

we use this and replace sin²(x) with 1-cos²(x)

4(1-cos²(x))+8cos(x) = 4-4cos²(x)+8cos(x)

-4cos²(x)+8cos(x)+4 = 7

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

4t²-8t-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

2. 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 $f^{-1}(x)$

3. yes i know that i meant that the wrong is in x1=pi - pi/3 + n *2pi

4. i think maybe it will be x1=pi/3 + n *2pi or x1=pi - pi/3 - n *2pi

right now ?

5. 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.