Can only prove 2 out of 4 solutions...wouldnt mind be pointed in the right direction

frank789

Junior Member
Joined
Sep 16, 2017
Messages
58
Hi All,

Having a bit of trouble with this one, I can find 2 solutions but the answer key is calling for 4.

Find all values of x in the interval [0,2pi] that satisfy

sin2x=cosx

heres how I went about it

sin2x = cosx
2sinxcosx = cosx

then cancel the cos and divide

sinx = 1/2

which gives solutions of pi/6 and 2pi/3. Not sure where the other 2 solutions are coming from as the identity rearranges to this and sin only gives positive answers in 2 quadrants however the answer key is giving me an answer for all 4 quadrants.

dont want the answer but a nudge in the right direction would be super
 
Hi All,

Having a bit of trouble with this one, I can find 2 solutions but the answer key is calling for 4.

Find all values of x in the interval [0,2pi] that satisfy

sin2x=cosx

heres how I went about it

sin2x = cosx
2sinxcosx = cosx → cos(x) * [2sin(x) - 1] = 0

sinx = 1/2 or cos(x) = 0

then cancel the cos and divide

sinx = 1/2

which gives solutions of pi/6 and 2pi/3. Not sure where the other 2 solutions are coming from as the identity rearranges to this and sin only gives positive answers in 2 quadrants however the answer key is giving me an answer for all 4 quadrants.

dont want the answer but a nudge in the right direction would be super
.
 

ok I follow

you essentially saying 0 = 0 so since cos has 0 at pi/2 and sin 2x or (sin2(pi/2) is 0 then it also satisfies the equation

i hope you dont mind me extending my original question then to is there a way to tell how many solutions you are going to get from a particular equation by looking at it?
 
That depends on what you mean by "looking at it". When working with periodic functions, ALWAYS assume many, many solutions until something tells you otherwise.
 
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That depend on what you mean by "looking at it". When working with periodic functions, ALWAYS assume many, many solutions until something tells you otherwise.

ok well let me rephrase

how do you know to stop looking for solutions?
 
ok well let me rephrase

how do you know to stop looking for solutions?
Here you had a restricted domain of 0≤x≤2π
As mmm responded, there are infinite solutions. But there are 4 solutions within the given domain.
As soon as your answers stray beyond the given domain - stop looking.
 
When you said "cancel the cos" you meant "divide both sides by cos(x). But you cannot divide by 0. Dividing both sides by something is only possible when that "something" is not 0.
 
When you said "cancel the cos" you meant "divide both sides by cos(x). But you cannot divide by 0. Dividing both sides by something is only possible when that "something" is not 0.
Welcome back ...... long time no hear (well may be "see" is more accurate)
 
When you said "cancel the cos" you meant "divide both sides by cos(x). But you cannot divide by 0. Dividing both sides by something is only possible when that "something" is not 0.

my apologizes for terminology shortcuts haha by cancel i meant reduce cosx/cosx to 1 in the identity
 
It's not a "terminology shortcut". It's an error.

Thought Question: Are you SURE that cos(x) / cos(x) = 1. What if cos(x) = 0?
 
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Find all values of x in the interval [0,2pi] that satisfy

sin2x=cosx

heres how I went about it

sin2x = cosx

2sinxcosx = cosx

then cancel the cos and divide

sinx = 1/2
By dividing each side by cos(x), you eliminated the possibility that cos(x)=0 because we cannot divide by zero.

That's why you missed those two solutions. Subhotosh showed how to avoid that. :cool:
 
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