How do you solve $3 cos^2 X + 7 sin X = 5$ ?

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Answer 1 · 2016-05-22T14:18:57

General solution of $3sin^2X-7sinX+2=0$ is

$npi+(-1)^npi/9.244$

$3cos^2X+7sinX=5$

or $3(1-sin^2X)+7sinX-5=0$

or $-3sin^2X+3+7sinX-5=0$

or $3sin^2X-7sinX+2=0$

or $3sin^2X-6sinX-sinX+2=0$

or $3sinX(sinX-2)-1(sinX-2)=0$

or $(3sinX-1)(sinX-2)=0$

But as range of sinX is within $[-1,1]$

$sinX-2!=0$ and hence

$3sinX-1=0$ or $sinX=1/3$ or $X=19.47^o=pi/9.244$

Hence General solution of $3sin^2X-7sinX+2=0$ is

$npi+(-1)^npi/9.244$

How do you solve 3 cos^2 X + 7 sin X = 53 cos^2 X + 7 sin X = 5?