Question

PLEASE HELP
(No calculator) Find all exact solutions to the equation sin^2x+ 2cos^2x= -2 with x between [0, 2pi]. You have an intermediate
step in which you find solutions to two dierent linear trigonometric equations. Identify the two equations that you used.

Answer 1

trick is to rewrite
\[\sin^2(x)=1-\cos^2(x)\]

Answer 2

then you will have a quadratic equation in cosine

Answer 3

lets just call cosine \[a\] to make the algebra easier

Answer 4

\[1-a^2+2s^2=-2\]
\[1+a^2=-2\]
\[a^2=-3\] and either i have made a mistake, or there is no solution, or perhaps you typed it in wrong

Answer 5

oh for the love of pete. the left hand side is the sum of two squares. the right hand side is negative. no solution by inspections

Answer 6

sure it wasn't
\[sin^2(x)+2cos^2(x)=2\]?

Answer 7

Oh how confusing. These are my available answers:

A) cos x = 0; tan x = 1
B) sin x = -1/2; sin x = 2
C) sin x = 0; tan x = -1
D) cos x = -1/2; cos x = 1/2
E) cos x = -1; cos x = 3

Answer 8

wait wait. are you sure it is what i wrote?

Answer 9

because no matter what x is
\[sin^2(x)\geq0\] and likewise
\[cos^2(x)\geq 0\] so there is not way to add them and get -2 !

Answer 10

it definitely says -2 :(

\[\sin^2 x + 2cosx =-2\]

Oh gosh I am really sorry the cosine was not squared! My mistake!

Answer 11

oooooooooooooooooh k

Answer 12

now we proceed as before. rewrite as
\[1-\cos^2(x)+2\cos(x)=-2\]

Answer 13

clear?

Answer 14

yes

Answer 15

then we get a quadratic equation in cosine
\[-cos^2(x)+2cos(x)+3=0\] or
\[cos^2(x)-2cos(x)-3=0\]

Answer 16

like solving
\[a^2-2a-3=0\]
\[(a-3)(a+1)=0\]
\[a = 3\] or \[a=-1\]

Answer 17

now back to trig:
\[cos(x)=3\] well that is impossible since the biggest cosine can be is 1

Answer 18

\[cos(x)=-1\] which means
\[x=\pi\]

Answer 19

How do I use x= Pi to choose one of the possible answers?

Answer 20

let me look at your answers
A) cos x = 0; tan x = 1
B) sin x = -1/2; sin x = 2
C) sin x = 0; tan x = -1
D) cos x = -1/2; cos x = 1/2
E) cos x = -1; cos x = 3

Answer 21

we started with \[sin^2(x)+2cos(x)=−2\] yes?

Answer 22

number 1 is out since if cos(x)=0 there is no was sin^2(x)=-2
number 2 is out since sin(x) cannot be 2

man i am lost lost lost. maybe they just want you to write
E) because that is what we got. but this doesn't make sense because there is no was cos(x) = 3

Answer 23

ooh i see. it just asks for the "intermediate step" and that is what we got!

Answer 24

we got E) as an "intermediate step" not clearly the final step

Answer 25

the final step is to say that cos(x)=3 is impossible so
cos(x)= -1 so
\[x=\pi\]

Answer 26

sorry but this question is confusing because you are asked to write something that is not possible. but ok it is an 'intermediate step'

Answer 27

I get it! Ok so heres a douzy. The last question is written EXACTLY the same, but apparently I am supposed to solve for it a different way because the new possible answers are :
A) cos x = -1; cos x = 3
B) sin x = 0; cos x = -1/2
C) cos x = 0; tan x = 1
D) sin x = -1/2; sin x = 2
E) cos x = -1/2; cos x = 1/2

Answer 28

but answer A) is identical to answer E from the previous problem!

Answer 29

where are you getting these?

Answer 30

Uggghhhh maybe they want me to use the formula for 2cosx?

its a worksheet :(

Answer 31

now it is answer A) which as you can see is answer E) from before. you cannot do this a different way

Answer 32

5 and 6 are the same. same problem, same answer. ignore it

Answer 33

Ok, thank you!

Answer 34

What is your opinion on
Find all solutions of :
sqrt{3}tan(x) +1 = 0

I know the solution is x = -1/sqrt{3}, and I know tangent of POSITIVE 1/sqrt{3} is pi/6.......but what does the negative do to the answer?
As far as I know the solution is
x= pi/6 + pik
x= 5pi/6 +pik