Solve cos(4x)-cos(2x)=0 ∀ 0<=x<=2pi ..............(0)
Normal solution:
1. use the double angle formula to decompose, and recall cos^2(x)+sin^2(x)=1
cos(4x)=cos^2(2x)-sin^2(2x)=2cos^2(2x)-1 .................(1)
2. substitute (1) in (0)
2cos^2(2x)-1-cos(2x)=0
3. substitute u=cos(2x)
2u^2-u-1=0
4. Solve for x
factor
(u-1)(u+1/2)=0
=> u=1 or u=-1/2
However, since cos(x) is an even function, so solutions to
{cos(2x)=1, cos(-2x)=1, cos(2x)=-1/2 and cos(-2x)} ...........(2)
are all solutions.
5. The cosine function is symmetrical about pi, therefore
cos(-2x)=cos(2*pi-2x),
solution (2) above becomes
{cos(2x)=1, cos(2pi-2x)=1, cos(2x)=-1/2, cos(2pi-2x)=-1/2}
6. Solve each case
cos(2x)=1 => x=0
cos(2pi-2x)=1 => cos(2pi-0)=1 => x=pi
cos(2x)=-1/2 => 2x=2pi/3 or 2x=4pi/3 => x=pi/3 or 2pi/3
cos(2pi-2x)=-1/2 => 2pi-2x=2pi/3 or 2pi-2x=4pi/3 => x=2pi/3 or x=4pi/3
Summing up,
x={0,pi/3, 2pi/3, pi, 4pi/3}
Answer:
C. 3x + 2y = 5 and 2y = 6 - 3x & D. 3x = 2 − 4y and 4y = 7 − 3x , have no solution! (:
Answer:
<em>x</em> = -6; <em>y</em> = 2 is the solution of the given system of equations.
Step-by-step explanation:
The given equations are :
<em>y</em> = 5<em>x</em> + 32 .......(1)
<em>y</em> = -4<em>x</em> - 22 .......(2)
Substituting the value of '<em>y'</em> from equation (1) in equation (2), we get
5<em>x</em> + 32 = -4<em>x</em> - 22
⇒5<em>x</em> + 4<em>x</em> = -22 - 32
⇒9<em>x</em> = -54
⇒<em>x</em> = (-54) ÷ 9
⇒<em>x</em> = -6
Put <em>x</em> = -6 in equation (1), we get
<em>y</em> = 5 × (-6) + 32 = -30 + 32 = 2
So, <em>x</em> = -6; <em>y</em> = 2 is the solution of the given system of equations.
Answer:
Maxon is!
Step-by-step explanation:
Slope is the change in y over the change in x is which is what Maxon is saying.
