Question

cos2x- sqrt 2 sinx=1 Find all solutions

Answer 1

well \[\cos(2x) = 1 - 2\sin^2(x) \]

so make the substitution and you'll have a quadratic equation

\[1 - 2\sin^2(x) -\sqrt{2}\sin(x)= 1\]

rewriting the equation and collecting like terms you get

\[\sin^2(x) + \sqrt{2} \sin(x) = 0\]

this can now be solved by factoring

Answer 2

@campbell_st where did the sin^2 come from?

Answer 3

ok... quick question is it

1. cos(2x)

or

2. \[\cos^2(x)\]

Answer 4

1. cos2x (there is no exponents)

Answer 5

ok... great, the stuff above is correct.

cos(2x) can be rewritten is several forms, and they are known facts about cos(2x)

to show you if you are familar with the sum of 2 angles

cos(x + x) = cos(x) cos(x) - sin(x)sin(x)
= cos^2(x) - sin^2(x) (equation 1)

now you may know sin^2(x) + cos^2(x) = 1

by making cos^2(x) the subject I get cos^2(x) = 1 - sin^2(x) (equation 2)

if I substitute equation 2 into equation 1 you get 1- 2sin^2(x)

its just a property you'll need to know....

so, long story short... I made the substitution

cos(2x) = 1 - sin^2(x)

which means you now have a "quadratic" equation in sin(x) that can be solved for x.

sorry to be long winded but I hope it makes sense.

Answer 6

oops missing a 2

cos(2x) = 1 - 2sin^2(x)

Answer 7

umm.. so cosx= pi/3

Answer 8

no you have

\[2\sin^2(x) + \sqrt{2} \sin(x) = 0\]

take sin(x) as out as a common factor and you get

\[\sin(x)(2\sin(x) + \sqrt{2}) = 0\]

so you are looking

where does

sin(x) = 0

and where does

\[2\sin(2) + \sqrt{2} = 0\]

these will be the solutions...

Answer 9

so sinx= 0 +2pi..etc and sinx=5pi/4+2pi

Answer 10

ummm \[\sin(x) = -\frac{\sqrt{2}}{2} \]

3rd quadrant is \[\frac{5\pi}{4}\]

and 4th quadrant

\[\frac{7\pi}{4}\]

Answer 11

hope that makes sense...

Answer 12

yes
its does i get that but has to be in terms or radians

Answer 13

you just need the 3rd angle in general solution form

Answer 14

ok i think i got it, thanks.

Answer 15

well you can either use

\[\frac{7\pi}{4} + 2\pi\]

Answer 16

yeah i just put all of them