Question #1a6f6

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Answer 1 · 2017-03-04T02:10:22

$5 cos^2(x)+ 9 sin(x) =9$

$=>5 -5 sin^2(x)+ 9 sin(x) =9$

$=>5 sin^2(x)- 9 sin(x) +9-5=0$

$=>5 sin^2x- 5sinx-4sinx +4=0$

$=>5 sinx(sinx-1)-4(sinx-1)=0$

$=>(sinx-1)(5 sinx-4)=0$

When

$sinx-1=0=>sinx=1=sin(pi/2)$

$=>x=npi+(-1)^npi/2" where " n in ZZ$

Again when

$(5 sinx-4)=0$

$=>sinx =4/5=sin(sin^-1(4/5))$

$=>x=npi+(-1)^nsin^-1(4/5)" where " n in ZZ$

where $sin^-1(4/5)~~0.3pi$