Question

resolva as equacoes a seguir
a) sen^2x=1
b) 2sen^2x+senx-1=0

Answer 1

a)

$sen~2x=1\\\\\boxed{2x\to y}\\\\sen~y=1\\\\y=arcsen~1\\\\y=\dfrac{\pi}{2}\\\\2x=\dfrac{\pi}{2}\\\\\boxed{\boxed{x=\dfrac{\pi}{4}}}$

b)

$2\cdot sen^2~x+sen~x-1=0\\\\\boxed{sen~x\to y}\\\\2y^2+y-1=0\\\\\Delta=b^2-4ac\\\\\Delta=1^2-4\cdot2\cdot(-1)\\\\\Delta=1+8\\\\\Delta=9\\\\y=\dfrac{-b\pm\sqrt\Delta}{2a}\\\\y=\dfrac{-1\pm3}{4}\\\\\begin{matrix}y'=\dfrac{1}{2}&y"=-1\end{matrix}\\\\\boxed{sen~x=y}\\\\\begin{matrix}sen~x'=\dfrac{1}{2}&sen~x''=-1\\\\x'=arcsen\Big(\dfrac{1}{2}\Big)&x''=arcsen(-1)\\\\\boxed{x'=\dfrac{\pi}{6}}&\boxed{x''=\dfrac{5\pi}{6}}&\boxed{x'''=\dfrac{3\pi}{2}}\end{matrix}$