cos4x = cos2x
We know that:
cos2x = 1-2cos^2 x
==> cos4x = 1-2cos^2 (2x)
Now substitute:
==> 1-2cos^2 (2x) = cos2x
==> 2cos^2 (2x) + cos2x - 1 = 0
Now factor:
==> (2cos2x -1)(cos2x + 1) = 0
==> 2cos2x -1 = 0 ==> cos2x =1/2 ==> 2x= pi/3
==> x1= pi/6 , 7pi/6
==> x1= pi/6 + 2npi
==> x2= 7pi/6 + 2npi
==> cos2x = -1 ==> 2x= pi ==> x3 = pi/2 + 2npi.
<span>==> x= { pi/6+2npi, 7pi/6+2npi, pi/2+2npi}</span>
Start at +1 on the x axis and then use 3/1 for rise over run. So you will start at +1 and then from that point rise 3 run 1
True just add two each time
Answer:
The third side is sqrt(77)
Step-by-step explanation:
Since this is a right triangle we can use the Pythagorean theorem
a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse
2^2 + b^2 = 9^2
4+ b^2 = 81
Subtract 4 from each side
b^2 = 81-4
b^2=77
Take the square root of each side
sqrt(b^2) = sqrt(77)
b = sqrt(77)
The third side is sqrt(77)
Answer:
q=-5
Step-by-step explanation:
6+q=1
Isolate the variable by subtracting 6 from both sides of the equation
q=1-6
q=-5
