Question

Va rog mult, am maine teza ..

Answer 1

1.

$sin^2 x + cos^2 x = 1$
$9/16 + cos^2x = 1$
[tex]cos^2x = 7/16 [/tex]
x ∈ Cadranului I => cos x > 0
$cos x = \sqrt{7} /4$

3.

Notam cos x  = y
Echivalent cu
3y² + 5y + 2 = 0
Δ = 25 - 12*2 = 1
y₁= $\frac{-5 + 1}{6} = -2/3$
y₂ = $\frac{-5 - 1}{6} = -1$

=> x₁ = arccos(-2/3)
=> x₂ = arccos(-1) = π