Question

Dacă tgx+ctgx=2, să se calculeze sin2x și cos2x.

Answer 1

tgx=t
t+1/t=2
t²-2t=1=0
(t-1)²=0
t=1
tgx=1
x=π/4+kπ

2x=π/2+2kπ
sin2x=1
cos2x=0

as simple as that!

Answer 2

[tex]\it tgx+ctgx=2 \Rightarrow \dfrac{^{sinx)}sinx}{\ \ cosx} + \dfrac{^{cosx)}cosx}{\ \ sinx} =\ ^{sinx cosx)}2 \Rightarrow \\ \\ \\ \Rightarrow sin^2x+cos^2x=2sinxcosx \Rightarrow 1 =sin2x \Rightarrow sin2x=1 \\ \\ \\ sin2x=1 \Rightarrow sin^2 2x =1 \Rightarrow 0 = 1-sin^2 2x \Rightarrow cos^2 2x=0 \Rightarrow \\ \\ \\ \Rightarrow cos2x=0[/tex]