Question

sin²θ-cos²θ-cosθ=1. Find all solutions in the interval [0,2π).

Answer 1

\[sin^2(x)-cos^2(x)-cos(x)=1\]
first rewrite in term of cosine alone, using the identity
\[sin^2(x)+cos^2(x)=1\]
\[1-cos^2(x)-cos^2(x)-cos(x)=1\]
\[1-2cos^2(x)-cos(x)=1\]
\[-2cos^2(x)-cos(x)=0\]
\[2cos^(x)+cos(x)=0\]
\[cos(x)(2cos(x)+1)=\]
\[cos(x) = 0\]
or
\[2cos(x)+1=0\]

Answer 2

of course you still have to solve
\[cos(x) = 0\]
\[x=\frac{\pi}{2}\]or \[x=\frac{3\pi}{2}\]

Answer 3

\[2cos(x)+1=0\]
\[cos(x)=-\frac{1}{2}\]
\[x=\frac{2\pi}{3}\]
\[x=\frac{4\pi}{3}\]

Answer 4

so, x=2π/3 and 4π/3?

Answer 5

you have 4 solutions all together.

Answer 6

oh, ok...also the π/2 and 3π/2....
thanks

Answer 7

right. is it clear?

Answer 8

it is, when someone shows you!

Answer 9

mmm rly u have unfintly solutin here
by adding the period of ur functin

Answer 10

of course!
idea is to write using one trig function, then you get a quadratic equation to solve.
there are an infinite number of solutions, that is why the question restricts interval from
\[[0,2\pi)\]

Answer 11

i.e ur sol + 2n bay

Answer 12

thanks 'satellite 73' . hard to know where to start.

Answer 13

welcome!

Answer 14

mm sry 'm jst dont care 2 the question

Answer 15

it is, when someone shows you!

Answer 16

oh, ok...also the π/2 and 3π/2....
thanks

Answer 17

so, x=2π/3 and 4π/3?

Answer 18

yes 4 total.
two for the equation
\[cos(x)=0\]
and two for \[cos(x)=-\frac{1}{2}\]