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 }
7=thousands
8=hundreds
5=tens
4=ones
2=tenths
0=hundredths
9=thousandths
Solving for a gives us: 
Step-by-step explanation:
Solving an equation for a variable means that we have to isolate the variable in the equation
Given equation is:

For isolating a, subtracting c from both sides

Subtracting b from both sides

Hence,
Solving for a gives us: 
Keywords: Equation, formula
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9514 1404 393
Answer:
Step-by-step explanation:
The area of a square is the square of the side length. Then the side length is the square root of the area.
- 6² = 36 . . . area of the square of leg 6
- 8² = 64 . . . area of the square of leg 8
- 36+64 = 100 . . . sum of the two leg squares = area of hypotenuse square
- √100 = 10 . . . hypotenuse length