Solve the equation on the interval x is greater than 0 an less than 2pi sin(2x-pi/2)=1 Please show step by step instructions.
You should solve the equation using the followings such that:
`sin x = a => x = (-1)^n arcsin a + npi`
Reasoning by analogy yields:
`sin(2x-pi/2)=1 => 2x-pi/2 = (-1)^n arcsin 1 + npi`
`2x - pi/2 = (-1)^n*pi/2 + npi`
`2x = (-1)^n*pi/2 + npi + pi/2`
`x = (-1)^n*pi/4 + npi/2 + pi/4`
For `n = 1 => x = -pi/4 + pi/2 + pi/4 = pi/2`
For `n = 2 => x = pi/4 + pi + pi/4 => x = pi/2 + pi = 3pi/2`
For `n = 3 =>x = -pi/4 + 3pi/2 + pi/4 = 3pi/2`
` n = 4 => x = pi/4 + 2pi + pi/4`
Notice that for `n = 4 => x = 5pi/4 > 2pi`
Hence, evaluating the solutions to the given equation, in `(0,2pi),` yields `x = pi/2` and `x = 3pi/2` .
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