How do you solve $4sin^2x+5=6$ ?

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Answer 1 · 2016-09-27T17:09:05

$x=npi+-pi/6$ , where $n$ is an integer

$4sin^2x+5=6$

$hArr4sin^x+5-6=0$ or

$4sin^x-1=0$

or $(2sinx)^2-1^2=0$

or $(2sinx+1)(2sinx-1)=0$

so either $2sinx=-1$ or $2sinx=1$

i.e. $sinx=-1/2$ or $sinx=1/2$

Now $sinx=1/2$ , when $x=pi/6$ or $(5pi)/6$

and $sinx=-1/2$ , when $x=-pi/6$ or $-(5pi)/6$

as $(5pi)/6=pi-pi/6$ , we can summarize result, in general, as

$x=npi+-pi/6$ , where $n$ is an integer

How do you solve 4sin^2x+5=64sin^2x+5=6?