Question

Please help will medal

http://prntscr.com/anzdnc

Answer 1

do you know the general formula for \(tan\) and \(sin\) ?

Answer 2

yes

Answer 3

okay
we simplify the 1st one->

\(tan^2 x+tanx=0\)
\(tanx(tanx+1)=0\)

so either \(tanx=0\)
or \(tanx+1=0\) -> \(tanx=-1\)

now try finding the solutions

Answer 4

y is it tanx+1?

Answer 5

i dont get this

Answer 6

@phi

Answer 7

forget about tan for the moment

if you had
y^2 + y= 0
or , after "factoring out" they y:
y(y+1) = 0

you have two terms: y and (y+1)
if you multiply two things and get 0
then you say: one of them (or maybe both!) must be 0
for example, if y=0 you would have 0 * (0+1) = 0*1 and that is zero
or if y=-1, you would have -1* (-1+1) = -1* 0 and that is zero

Answer 8

you are beast

Answer 9

use that idea for your problem
\[ \tan^2 x + \tan x = 0 \]

Answer 10

so if we factor tan x then this is equal to what @imqwerty said right?

Answer 11

tanx(tanx+1)=0

Answer 12

yes, and to get 0 , one or the other of the terms must be zero
so one solution is
tan x = 0
and the other solution is
tan x + 1 = 0

Answer 13

ok, what about the second one?

Answer 14

they restrict the answer to be between -pi/2 and pi/2 (-90 deg and + 90 degrees)

tan x + 1 = 0
add -1 to both sides, and simplify:
tan x + 1 - 1= 0 -1
tan x = -1

you should know what angle "x" is (you memorized sin, cos, tan values for 0, 30, 45, 60 and 90 degrees, right?)

Answer 15

or you know tan A = y/x