How do you solve for 0 ≤ x < 2π using the equation 4sin ² x - 1 = 0 ?

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Answer 1 · 2018-02-28T17:51:17

$pi/6,(5pi)/6,(7pi)/6,(11pi)/6$

$4sin^2x-1=0rArrsin^2x=1/4rArrsinx=color(red)(+-1/2)$
Given, $0<=x<2pi$

$0<=x<(pi/2)rArrsinx=1/2=sin(pi/6)rArrx=color(red)(pi/6)$
$(pi/2) <= x<(pi)rArrsinx=1/2=sin(pi-pi/6)=sin((5pi)/6)rArrx=color(red)((5pi)/6)$
$pi<=x<(3pi)/2rArrsinx=-1/2=sin(pi+pi/6)=sin((7pi)/6)rArrx=color(red)((7pi)/6)$
$(3pi)/2<=x<2pirArrsinx=-1/2=sin(2pi-pi/6)=sin((11pi)/6)rArrx=color(red)((11pi)/6)$
Note:The general solution is
$sinx=+-1/2rArrsinx=sin(+-pi/6)$
So,
$x=kpi+(-1)^k(+-pi/6),kinZ$
Taking $k=0,+-1,+-2,...$ we can obtain above answer.