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sin^2(x) – 3sin(x) + 2 = 0 within [-4π, 4π]
Solution
Let y = sin(x)
Then the given equation becomes
y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0
y = 1 (Note that 2 cannot be a value for a sine function)
x = arcsin(1) = π/2
sin^2(x) – 3sin(x) + 2 = 0 within [-4π, 4π]
Solution
Let y = sin(x)
Then the given equation becomes
y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0
y = 1 (Note that 2 cannot be a value for a sine function)
x = arcsin(1) = π/2