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I knew that the root was θ = -78.07, but for some reason that did not give the correct solution when you sub it back into 3cosx=2sin(x+60) for some reason ..

but I noticed θ = 78.07 being a root, so I stated that.

It is weird, not to get θ = 78.07° graphically ...

Ricki

I knew that the root was ? = -78.07, but for some reason that did not give the correct solution when you sub it back into 3cosx=2sin(x+60) for some reason ..

but I noticed ? = 78.07 being a root, so I stated that.

It is weird, not to get ? = 78.07° graphically ...

but I noticed ? = 78.07 being a root, so I stated that.

It is weird, not to get ? = 78.07° graphically ...

I think there is a problem with squaring both sides, because when you square root later on you introduce extra illegal solutions. I think we have to use a different method. Like Rsin(theta + alpha)

steve2005

I think there is a problem with squaring both sides, because when you square root later on you introduce extra illegal solutions. I think we have to use a different method. Like Rsin(theta + alpha)

Ah right! that must be true, its always you have to be careful when squaring.

Well, my method does partially work, as the second solution was just θ = 360 - Ø where Ø was ≈ 78.07.

About putting it in the form Rsin(θ+Ø) , can you do this and show me, as I'm not familar with this yet!

Ricki

3cosx=2sin(x+60)

3cosx=2(sinxcos60+cosxsin60)

3cosx=2(sinx/2+√3cosx/2)

3cosx=sinx+√3cosx

9cos²x=(sinx+√3cosx)²

9cos²x = sin²x+2√3cosxsinx + 3cos²x

9cos²x = 1 - cos²x + 2√3cosxsinx + 3cos²x

7cos²x = 1 + 2√3cosxsinx

7 = 1/cos²x + 2√3sinx/cosx

7 = sec²x + 2√3tanx

Using sec²x = tan²x + 1 ...

7 = (tan²x + 1) + 2√3tanx

tan²x + 2√3tanx - 6 =0

sub t = tanx

t² + 2√3t - 6 = 0

(t + √3)² - 3 - 6 = 0

t = -√3 ±√9

sub t = -√3 + √9 into t = tanx ...

arctan(-√3 + √9) = θ

θ ≈ 51.74° etc.

sub t = -√3 - √9 into t = tanx ...

arctan(-√3 - √9) = θ

θ ≈ 78.07° etc.

3cosx=2(sinxcos60+cosxsin60)

3cosx=2(sinx/2+√3cosx/2)

3cosx=sinx+√3cosx

9cos²x=(sinx+√3cosx)²

9cos²x = sin²x+2√3cosxsinx + 3cos²x

9cos²x = 1 - cos²x + 2√3cosxsinx + 3cos²x

7cos²x = 1 + 2√3cosxsinx

7 = 1/cos²x + 2√3sinx/cosx

7 = sec²x + 2√3tanx

Using sec²x = tan²x + 1 ...

7 = (tan²x + 1) + 2√3tanx

tan²x + 2√3tanx - 6 =0

sub t = tanx

t² + 2√3t - 6 = 0

(t + √3)² - 3 - 6 = 0

t = -√3 ±√9

sub t = -√3 + √9 into t = tanx ...

arctan(-√3 + √9) = θ

θ ≈ 51.74° etc.

sub t = -√3 - √9 into t = tanx ...

arctan(-√3 - √9) = θ

θ ≈ 78.07° etc.

Yeah, the book gives 51.7 and 231.7 as the answers.

mala2k

We should all be ashamed of ourselves here , we over compicated this question:

3cosx=2sin(x+60)

3cosx=2(sinxcos60+cosxsin60)

3cosx=2(sinx/2+?3cosx/2)

3cosx=sinx+?3cosx

Instead of squaring divide by cosx:

3=tanx+?3

tanx=3-?3

x=51.7

x=180+51.7=231.7

3cosx=2sin(x+60)

3cosx=2(sinxcos60+cosxsin60)

3cosx=2(sinx/2+?3cosx/2)

3cosx=sinx+?3cosx

Instead of squaring divide by cosx:

3=tanx+?3

tanx=3-?3

x=51.7

x=180+51.7=231.7

Talk about over complicated...... this is using Rcos(x + alpha) ..

mala2k

We should all be ashamed of ourselves here , we over compicated this question:

3cosx=2sin(x+60)

3cosx=2(sinxcos60+cosxsin60)

3cosx=2(sinx/2+√3cosx/2)

3cosx=sinx+√3cosx

Instead of squaring divide by cosx:

3=tanx+√3

tanx=3-√3

x=51.7

x=180+51.7=231.7

3cosx=2sin(x+60)

3cosx=2(sinxcos60+cosxsin60)

3cosx=2(sinx/2+√3cosx/2)

3cosx=sinx+√3cosx

Instead of squaring divide by cosx:

3=tanx+√3

tanx=3-√3

x=51.7

x=180+51.7=231.7

Oh my god, that would've been so much easier!

mala2k

We should all be ashamed of ourselves here , we over compicated this question:

3cosx=2sin(x+60)

3cosx=2(sinxcos60+cosxsin60)

3cosx=2(sinx/2+√3cosx/2)

3cosx=sinx+√3cosx

Instead of squaring divide by cosx:

3=tanx+√3

tanx=3-√3

x=51.7

x=180+51.7=231.7

3cosx=2sin(x+60)

3cosx=2(sinxcos60+cosxsin60)

3cosx=2(sinx/2+√3cosx/2)

3cosx=sinx+√3cosx

Instead of squaring divide by cosx:

3=tanx+√3

tanx=3-√3

x=51.7

x=180+51.7=231.7

I have my reason, of being in a hurry, so was rushing things!

Nice elegant solution.

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