How do you solve $4sin^2 theta - 13sin theta +3 = 0$ ?

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Answer 1 · 2015-04-30T14:43:54

In this way:

let $t=sinx$ ,

so:

$4t^2-13t+3=0$

$Delta=b^2-4ac=169-48=121=11^2$

$t_(1,2)=(-b+-sqrtDelta)/(2a)=(13+-11)/8rArr$

$t_1=(13+11)/8=3$

$t_2=(13-11)/8=2/8=1/4$ .

And now:

$sinx=3$ impossible because the range of the sinus function is $[-1,1]$ ,

$sinx=1/4rArr$

$x=arcsin(1/4)+2kpi$

$x=pi-arcsin(1/4)+2kpi$ .

This is because the solutions of an equation in the variable sinus has (when it has!) 2 supplementary solutions.

How do you solve 4sin^2 theta - 13sin theta +3 = 04sin^2 theta - 13sin theta +3 = 0?