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sinx−1=0 or 2sinx+1=0
sinx=1 or 2sinx=−1
x=2π+π2, where is any integer
sinx is negative in quadrant 3 and 4.
Then general solution is
⇒x=7π6+2kπ,11π6+2kπ for any inter
In 0≤x<2π, put x=0 we get
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