Question

trig help..
solve each equation for solutions in the interval [0, 360)
tan θ +1 = √3 + √3cotθ?
any help would be appreciated!

Answer 1

$$
tan(\theta)+1 = \sqrt{3}+\sqrt{3}cot(\theta)\\
tan(\theta)+1 = \sqrt{3}(1+cot(\theta))\\
\cfrac{tan(\theta)+1}{1+cot(\theta)} = \sqrt{3}
$$

Answer 2

so, from the trig identities
tan = sin/cos
cot = cos/sin

we'll expand that to

Answer 3

$$
\cfrac{
\cfrac{sin(\theta)}{cos(\theta)}+1}
{
1+\cfrac{cos(\theta)}{sin(\theta)}}\\
\implies
\cfrac{
\cfrac{\color{blue}{sin(\theta)+cos(\theta)}}
{cos(\theta)}}
{
\cfrac{\color{blue}{sin(\theta)+cos(\theta)}}
{sin(\theta)}}\\
$$

Answer 4

well all that
\(\huge = \sqrt{3}\)

Answer 5

see any like terms to cancel out?

Answer 6

yes, but where did the +1's go?

Answer 7

in the fraction addition

s/c +1
GCF of both fractions, is c
so
(s+c)/c

Answer 8

you're really adding
sine/cosine + 1/1

Answer 9

okay, thanks! i get it now :)

Answer 10

$$
\cfrac{
\cfrac{\color{blue}{sin(\theta)+cos(\theta)}}
{cos(\theta)}}
{
\cfrac{\color{blue}{sin(\theta)+cos(\theta)}}
{sin(\theta)}}= \sqrt{3}\\
\implies tan(\theta) = \sqrt{3}
$$

Answer 11

now you just need to find in your Unit Circle, what angle has a tangent of \(\sqrt{3}\)
or just arctan both sides

Answer 12

okay, thanks for your time :)

Answer 13

yw

Answer 14

@jdoe0001 so the answer would be π/3 and 4π/3 right?

Answer 15

yes

Answer 16

thank you!

Answer 17

I and II quadrants, only place where tangent is positive

Answer 18

woops, I and III rather

Answer 19

that's what i got
thanks again!

Answer 20

yw