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October 30, 2014

October 30, 2014

Posted by **Anonymous** on Tuesday, May 22, 2012 at 11:27pm.

a) + or - 3/8pi +Kpi

b) 2/3pi + Kpi, 5/3pi + Kpi

c) + or - 2/5 +Kpi

d) 2/3 pi + 2Kpi, 5/3pi + 2Kpi

I really need help on this question :(

- trig -
**Reiny**, Wednesday, May 23, 2012 at 4:56amcos 2x + √2/2 =0

cos 2x = -√2/2

I know cos π/4 = √2/2 or cos 45° = +√2/2

but 2x could be in II or III

2x = π-π/4 or 2x = π + π/4

2x = 3π/4 or 2x = 5π/4

x = 3π/8 or x = 5π/8 ---> 67.5° or 112.5°

but the period of cos 2x = π

so adding/subtraction multiples of π to an answer will yield more answers

general solution

x = 3π/8 + kπ, x = 5π/8 + kπ

a) fits the first of my solutions, but does not include the 2nd part of the solution

the others are not correct.

BTW, you can illustrate my 2nd solution, which they don't have, on a calculator

e.g. let x = 112.5° + 17(180)° = 3172.5

LS = cos 2(3172.5) + √2/2

= -.707106781 + .707106781

= 0

= RS

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