Question #4bf05

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Answer 1 · 2017-02-02T08:22:32

The solutions are $S={-pi/6 , -5/6pi}$

We solve this equation like a quadratic equation

$ax^2+bx+c=0$

$4sin^2x+4sinx+1=0$

Let's calculate the discriminant

$Delta=b^2-4ac=4^4-4*4*1=16-16=0$

As $Delta =0$ , there is one double real root

$x=-b/2a$

$sinx=-4/(2*4)=-1/2$

Therefore,

$x=-pi/6 ; -5pi/6$

Answer 2 · 2017-02-02T15:47:13

$4sin^2x+4sinx+1=0$

$=>(2sinx)^2+2*2sinx*1+1^2=0$

$=>(2sinx+1)^2=0$

$=>sinx=-1/2=sin(-pi/6)$

General solution

$x=npi-(-1)^npi/6" where " n in ZZ$

We are to find out x for x $in [-pi;4pi]$

Putting $n=-1$

$x=-pi+pi/6=-(5pi)/6$

Putting $n=0$

$x=-pi/6 in [-pi;4pi]$

Putting $n=1$

$x=pi-(-pi/6)=(7pi)/6 in [-pi;4pi]$

Putting $n=2$

$x=2pi-pi/6=(11pi)/6 in[-pi;4pi]$

Putting $n=3$

$x=3pi-(-pi/6)=(19pi)/6 in [-pi;4pi]$

Putting $n=4$

$x=4pi-(+pi/6)=(23pi)/6 in [-pi;4pi]$