Question

a) 2.senx.COSX - COSX = 0
b) cos²x + senx = 1

Answer 1

item a)

$2\text{sen(x).cos(x)}-\text{cos(x)} = 0$

$\text{cos(x)}[\ 2\text{sen(x)}-1 \ ]= 0 \\\\ \text{cos(x)} = 0 \ \ \text{ou} \ \ 2.\text{sen(x)}-1 = 0$

então temos :

1)

$\displaystyle \text{cos(x)} = 0 \\\\\ \text x =\frac{\pi}{2}+2\text k.\pi \ \ ;\ \ \text x = \frac{3\pi}{2}+2\text k.\pi \ \ ,\boxed{\text{com k} \in \mathbb{Z}}$

2)

$\displaystyle 2\text{sen(x)}-1=0 \to \text{sen(x)} = \frac{1}{2} \\\\ \text x = \frac{\pi}{6}+ 2\text k.\pi \ \ ;\ \ \text x = \frac{5\pi}{6} + 2\text k.\pi \ \ ,\boxed{\text{com k} \in \mathbb{Z}}$

Soluções :

$\boxed{\text x = \{ \ \frac{\pi}{2}+2\text k.\pi \ \ ; \ \frac{3\pi}{2}+2\text k.\pi \ \ ; \ \frac{\pi}{6}+2\text k.\pi \ \ ; \ \frac{5\pi}{6}+2\text k.\pi \ \} \ ,\text k \in \mathbb{Z} }$

item b)

$\text{cos}^2(\text x)+\text{sen(x)}=1 \\\\ \text{1-sen}^2(\text x)+ \text{sen(x)} - 1 = 0 \\\\ \text{sen}^2(\text x)-\text{sen}(\text x) = 0 \\\\ \text{sen(x)}[\ \text{sen(x)} - 1 \ ] = 0 \\\\ \text{sen(x)} = 0 \ \ \text{ou} \ \ \text{sen(x)}-1 = 0 \to \text{sen(x)} = 1$

temos :

$\displaystyle \text{sen(x)} = 0 \\\\ \text x = 0 + 2\text k.\pi \to \boxed{\text x = 2\text k.\pi } \\\\ \boxed{\text x = \pi + 2\text k.\pi }$

e

$\displaystyle \text{sen(x)} = 1 \\\\ \boxed{\text x = \frac{\pi}{2}+2\text k.\pi } \\\\ \text$

Soluções :

$\boxed{\displaystyle \text x = \{ \ 2\text k.\pi \ ; \ \pi+2\text k.\pi \ ; \ \frac{\pi}{2} + 2\text k.\pi \ \} \ , \text{ k} \in \mathbb{Z}}$