Question

find all solutions in the internal (0.2pi]
cos 2x -square 2 sin-1 help

Answer 1

it appears that your answer will be a number set.
so let x=0.2 and solve, then
let x = pi and solve.
with the two solutions you get, these will be the number set representing the bounds of 'all solutions'

message me if you get stuck ^_^

Answer 2

i'm stuck :/

Answer 3

Help :/

Answer 4

(0.2pi] means (0.2, pi] right?

Answer 5

this
cos 2x -square 2 sin-1
means this
cos (2x) - sin^2 (-1)
right?

Answer 6

^_^ LOL

Answer 7

\[\cos2x - \sqrt{2} \sin-1\]

Answer 8

\[\Large \cos2x - \sqrt{2} \sin-1=0\]Recall your `Cosine Double Angle Identity`:\[\Large \color{#CC0033}{\cos2x\quad=\quad 1-2\sin^2x}\]

Answer 9

\[\Large 1-2\sin^2x-\sqrt2 \sin x-1=0\]Hmm looks like the 1's cancel out, that's convenient.
Do you need help from this point?

Answer 10

yes i'm

Answer 11

\[\Large \cancel1-2\sin^2x-\sqrt2 \sin x\cancel{-1}=0\]Multiply each side by -1 giving us,\[\Large 2\sin^2x+\sqrt2 \sin x\quad=\quad 0\]

Answer 12

Factor a sinx out of each term,\[\Large \sin x(2\sin x+\sqrt2)\quad=\quad 0\]To continue, set each factor equal to zero, then solve for sinx in each case,\[\Large \sin x=0\qquad\qquad\qquad 2\sin x+\sqrt2=0\]

Answer 13

Hmm you should get some nice special angles in each case.

Answer 14

like you

Answer 15

0, \[-\sqrt{2}/2\]

Answer 16

right