Do you want a general solution or from 0 <= x < 2pi?
tan^2(2x) - 1 = 0
tan^2(2x) = 1
Take the square root of both sides,
tan(2x) = +/- 1
Two equations:
tan(2x) = 1
tan(2x) = -1
Solve each equation.
tan(2x) = 1, 2x = {pi/4, 5pi/4, 9pi/4, 13pi/4},
x = { pi/8, 5pi/8, 9pi/8, 13pi/8 }
tan(2x) = -1, 2x = { 3pi/4, 7pi/4, 11pi/4, 15pi/4 },
x = { 3pi/8, 7pi/8, 11pi/8, 15pi/8}
So for solutions within [0, 2pi),
x = {pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8 }
Because the last equation already has the y eliminated so just need to get the other two to eliminate the y too
multiply 4 to the first equation to eliminate y, and you get
-4x-4y-4z=-32
-4x+4y+5z=7
now get rid of y so all you have left is
-8x+z=-25 (let call this equation #4)
now take equation #3 and #4 together to solve for one of the variables, i’ll solve for z first so i’ll get rid of x by multiply the equation #4 all by 4, you get
8x+8z=16
-8x+z=-25
solve it and get 9z= -9; z=-1
now you have z so take it and ply it into equation #3
2x+2(-1)=4
2x=6; x=3
take both of them, plug it in equation #1
-3-y-(-1)=-8
-y=-6 ; y=6
done so the answer is (x,y,z) = (3,6,-1)
Answer: first one (-4,1) and the second is (0,1)
Step-by-step explanation:
Well, they are ALIKE, but since a and b are different the terms are different. Hope this helped.
