How do you find the exact value of $5sin^2theta-4sintheta+cos2theta=0$ in the interval $0<=theta<2pi$ ?

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Answer 1 · 2017-04-06T13:06:00

$theta={0.334,pi/2,(pi-0.334)}$

$5sin^2theta-4sintheta+cos2theta=0$

$hArr5sin^2theta-4sintheta+1-2sin^2theta=0$

or $3sin^2theta-4sintheta+1=0$

and $sintheta=(-(-4)+-sqrt(4^2-4xx3xx1))/(2xx3)$

= $(4+-2)/6$ i.e. $sintheta=1=sin(pi/2)$

or $sintheta=1/3=sin0.334$

and $theta=2npi+pi/2$ or $npi+(-1)^n0.334$

and in $0 <= theta <= 2pi$

$theta={0.334,pi/2,(pi-0.334)}$

How do you find the exact value of 5sin^2theta-4sintheta+cos2theta=05sin^2theta-4sintheta+cos2theta=0 in the interval 0<=theta<2pi0<=theta<2pi?