Solve the following equation for all radian solutions and if 0\leq x\leq2\pi. Give all answers as exact values in radians. 0\leq x\leq2\pi

sanuluy 2021-09-12 Answered
Solve the following equation for all radian solutions and if 0x2π. Give all answers as exact values in radians.
0x2π
x=7π6,π2,11π6 rad
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Expert Answer

Tasneem Almond
Answered 2021-09-13 Author has 91 answers

Given equation:
2sin2xsinx1=0
2sin2x2sinx+sinx1=0
2sinx(sinx1)+1(sinx1)=0
(sinx1)(2sinx+1)=0
sinx1=0 or 2sinx+1=0
sinx=1 or 2sinx=1
sin=12
sinx=1
x=2π+π2, where is any integer
sinx is negative in quadrant 3 and 4.
Then general solution is
sinx=12
x=7π6+2kπ,11π6+2kπ for any inter
In 0x<2π, put x=0 we get
x=π2,7π6,11π6

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Jeffrey Jordon
Answered 2021-12-11 Author has 2495 answers

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