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Answer 1 · 2017-03-04T02:10:22

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

$\Rightarrow 5 - 5 {sin}^{2} (x) + 9 sin (x) = 9$ $=>5 -5 sin^2(x)+ 9 sin(x) =9$

$\Rightarrow 5 {sin}^{2} (x) - 9 sin (x) + 9 - 5 = 0$ $=>5 sin^2(x)- 9 sin(x) +9-5=0$

$\Rightarrow 5 {sin}^{2} x - 5 sin x - 4 sin x + 4 = 0$ $=>5 sin^2x- 5sinx-4sinx +4=0$

$\Rightarrow 5 sin x (sin x - 1) - 4 (sin x - 1) = 0$ $=>5 sinx(sinx-1)-4(sinx-1)=0$

$\Rightarrow (sin x - 1) (5 sin x - 4) = 0$ $=>(sinx-1)(5 sinx-4)=0$

When

$sin x - 1 = 0 \Rightarrow sin x = 1 = sin (\frac{π}{2})$ $sinx-1=0=>sinx=1=sin(pi/2)$

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

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