Question

2cosx-2secx=3

Solve for 0 less than or equal to x less than 2pi

Answer 1

\[\cos(x)-\frac{1}{\cos(x)}=\frac{3}{2}\] is a start

Answer 2

\[2cosx - 2secx = 3 Solve for 0\le x < 2pi\]

Answer 3

cosx=t
\[2t^2-3t-2=0\]

Answer 4

where did you get the squared from?

Answer 5

\[(t-2)*(2t+1)=0\]

Answer 6

\[t=2\quad and \quad t=-1/2\]

Answer 7

there is no cosx^2 in this problem, and then you cannot factor it out

Answer 8

(:

Answer 9

it is magic

Answer 10

haha what does that mean then?

Answer 11

trick is to put say \(\cos(x)=t\) as above and get
\[t-\frac{1}{t}=\frac{3}{2}\]
\[\frac{t^2-1}{t}=\frac{3}{2}\]
\[2(t^2-1)=3t\]
\[2t^2-2=3t\]
\[2t^2-3t-2=0\]
\[(t-2)(2t+1)=0\]
\[t=2,t=-\frac{1}{2}\]

Answer 12

then replace t by \(\cos(x)\)and get \(\cos(x)=2\) which is not possible or \(\cos(x)=-\frac{1}{2}\) which is possible

Answer 13

on the interval \((0,2\pi)\) we have
\[\cos(x)=-\frac{1}{2}\implies x=\frac{2\pi}{3}, x=\frac{4\pi}{3}\]

Answer 14

Thankyou! that was a great help