Thread: 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

You're working is fine - it's a domain and range issue - you can discard t2

The maxiumum value of is 1. Since then you cannot take the inverse cos.


It's not in the range of and because cos(x) and arccos(x) are inverses of each other then the range of becomes the domain of which means it's not in the domain of

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

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

right now ?

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.