Question

cos2x+ root3/2=0

Answer 1

\[2Tan^2x-Sec^2x+2tanx+2=0\]

Answer 2

\[\tan^2x+\tan^2x-\sec^2x+2tanx+2=0\]
\[\sec^2x-1=\tan^2x,~~~So~~~~\sec^2x-\tan^2x=1,~~~~substitute\]\[\tan^2x+(\tan^2x-\sec^2x)+2tanx+2=0\]
\[\tan^2x+1+2tanx+2=0\]\[\tan^2x+2tanx+3= 0\]

Answer 3

made an error in substituting, it is not 1, it's negative one.\[\tan^2x-\sec^2x=-1\]
So

Answer 4

\[\tan^2x+(-1)+2tanx+2=0\]\[\tan^2x+2tanx+1=0,~~~~~L e t~~tanx=a\]

Answer 5

\[a^2+2a+1=0~~~~~~~~~~~~~~~~~(a+1)^2=0\]

Answer 6

a=-1

Answer 7

\[tanx=-1\]

Answer 8

\[Tan ^{-1}(-1)=?\]

Answer 9

thank you so much! what is the answer in radians with a restriction of x is greater than or equal to 0. and x is less than or equal to 2pi

Answer 10

-pi / 4 i think.