All values of x in the interval [0, 2π] satisfy the equation. 7 sin(2x) = 7 cos(x) are (π/2), (3π/2), (π/6), and (5π/6).
An equation is formed when two equal expressions are equated together with the help of an equal sign '='.
The given equation can be solved for x as shown below,
7 sin(2x) = 7 cos(x)
7 sin(2x) - 7 cos(x) = 0
14 sin(x)cos(x) - 7 cos(x) = 0
[7cos(x) - 0] [2sin(x) - 1] = 0
Now, the factors can be written as,
[7cos(x) - 0] = 0
7 cos(x) = 0
cos(x) = 0
x = (π/2)+2πn ; (3π/2)+2πn
[2sin(x) - 1] = 0
2sin(x) = 1
sin(x) = 1/2
x = (π/6)+2πn ; (5π/6)+2πn
Combining all the solutions,
x = (π/2)+2πn, (3π/2)+2πn, (π/6)+2πn, (5π/6)+2πn, where n can be any whole number.
But since the solution is needed to be in the interval of [0, 2π], the value of x can be,
x = (π/2), (3π/2), (π/6), (5π/6)
Hence, All values of x in the interval [0, 2π] satisfy the equation. 7 sin(2x) = 7 cos(x) are (π/2), (3π/2), (π/6), and (5π/6).
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