làm hộ giúp mình vs
a)y=4cos(2x+pi/5)+9 .
b) y=1/√3cot2x+1
c) tanx-√3/2cosx+1=0.
d)√3tan3x+1=0
e)sin2x=√3-√3 cos2x
f)cos6x-3cos3x-1=0
g)cos4x-sin2x-1=0
i)2sin2x+1=0 với 0
làm hộ giúp mình vs a)y=4cos(2x+pi/5)+9 . b) y=1/√3cot2x+1 c) tanx-√3/2cosx+1=0. d)√3tan3x+1=0 e)sin2x=√3-√3 cos2x f)cos6x-3cos3x-1=0 g)cos4x-sin
By Kennedy
Giải thích các bước giải:
d.$\sqrt{3}.tan3x+1=0$
$\rightarrow tan 3x=\dfrac{-1}{\sqrt{3}}$
$\rightarrow tan 3x=tan \dfrac{-\pi}{6}$
$\rightarrow 3x=\dfrac{-\pi}{6}+k\pi$
$\rightarrow x=\dfrac{-\pi}{18}+\dfrac{k\pi}{3}$