Question #52f58

Question

1145 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-23T03:37:22

${cos}^{2} 3 x + 3 sin 3 x = 3$ $cos^2 3x+3sin3x=3$

$\Rightarrow 1 - {sin}^{2} 3 x + 3 sin 3 x = 3$ $=>1-sin^2 3x+3sin3x=3$

$\Rightarrow {sin}^{2} 3 x - 3 sin 3 x + 2 = 0$ $=>sin^2 3x-3sin3x+2=0$

$\Rightarrow {sin}^{2} 3 x - 2 sin 3 x - sin 3 x + 2 = 0$ $=>sin^2 3x-2sin3x-sin3x+2=0$

$\Rightarrow sin 3 x (sin 3 x - 2) - (sin 3 x - 2) = 0$ $=>sin3x(sin3x-2)-(sin3x-2)=0$

$\Rightarrow (sin 3 x - 2) (sin 3 x - 1) = 0$ $=>(sin3x-2)(sin3x-1)=0$

$sin 3 x = 2 \to not possible$ $sin3x=2->"not possible"$

So $sin 3 x = 1 = sin (\frac{π}{2})$ $sin3x=1=sin(pi/2)$

$=>3x=npi+(-1)^npi/2" where "ninZZ$

$=>x=(npi)/3+(-1)^npi/6" where "ninZZ$