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# Maths

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Solve the following equations for $$\theta$$, in the the interval $$0<\theta$$ (< or equals to 360). (Tbh i dont really know hot to insert the smaller sign with another line underneath it, hope u understood the the range correctly).

a) $$sin\theta + cos\theta =0$$

b) $$3 cos\theta = -2$$

c)$$(sin \theta -1)(5 cos \theta +3)=0$$

d) $$tan\theta = tan\theta(2+3 sin \theta )$$

Dec 27, 2017
edited by Rauhan  Dec 27, 2017

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Also... the LaTeX code is   \leq  .

Solve the following equations for  θ , in the the interval  0 < θ ≤  360° .

a)   sin θ + cos θ   =   0

Factor  cos θ  out of the terms on left side

(cos θ)( sin θ / cos θ + 1)   =   0

We can divide both sides by  cos θ  since  cos θ  can't be zero.

sin θ / cos θ  +  1   =   0

tan θ  +  1   =   0

tan θ   =   -1

And this occurs at     θ  =  135º     and     θ  =  315°

b)   3 cos θ   =   -2

Divide both sides by  3 .

cos θ   =   -2/3

Take the inverse cosine of both sides of the equation.

θ   =   arccos( -2/3 )

By putting this into a calculator, I get

θ   ≈   131.81°

And we know that the cosine is also negative in the third quadrant, so..

θ   ≈   360° - 131.81°   ≈   228.19°     also has a cosine of  -2/3 .

c)   (sin θ - 1)(5 cos θ + 3)   =   0

Here, we need to set each factor equal to zero and solve each for  θ  .

First factor:

sin θ - 1   =   0

sin θ   =   1         There is only one angle that has a sin of 1 .

θ   =   90°

Second factor:

5 cos θ + 3   =   0

5 cos θ   =   -3

cos θ   =   -3/5

θ   ≈   126.87°          and          θ   ≈   360° - 126.87°   ≈   233.13°

d)   tan θ   =   tan θ ( 2 + 3 sin θ )

Subtract  tan θ  from both sides.

0   =   tan θ ( 2 + 3 sin θ )  -  tan θ

Factor  tan θ  out of the terms on the right side.

0  =  (tan θ)( (2 + 3 sin θ)  -  1 )

0  =  (tan θ)( 1 + 3 sin θ)

Set each factor equal to zero and solve for  θ .

tan θ   =   0    This occurs at..

θ   =   180°     and     θ   =   360°

1 + 3 sin θ   =   0

3 sin θ   =   -1

sin θ   =   -1/3

θ   ≈   -19.47° + 360°   ≈   340.53°          and          θ   ≈   180° - -19.47°   ≈   199.47°

Dec 28, 2017
edited by hectictar  Dec 28, 2017

#1
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You can use this online Math keyboard to write the Math symbols you want:

http://math.typeit.org/

Dec 27, 2017
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If you're using a mac, you can do alt+< or > to get ≤ or ≥

MIRB14  Dec 27, 2017
edited by MIRB14  Dec 27, 2017
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I dont use mac but in the later question related to this chapter( if any) i will still try the command

Rauhan  Dec 28, 2017
#3
+8439
+3

Also... the LaTeX code is   \leq  .

Solve the following equations for  θ , in the the interval  0 < θ ≤  360° .

a)   sin θ + cos θ   =   0

Factor  cos θ  out of the terms on left side

(cos θ)( sin θ / cos θ + 1)   =   0

We can divide both sides by  cos θ  since  cos θ  can't be zero.

sin θ / cos θ  +  1   =   0

tan θ  +  1   =   0

tan θ   =   -1

And this occurs at     θ  =  135º     and     θ  =  315°

b)   3 cos θ   =   -2

Divide both sides by  3 .

cos θ   =   -2/3

Take the inverse cosine of both sides of the equation.

θ   =   arccos( -2/3 )

By putting this into a calculator, I get

θ   ≈   131.81°

And we know that the cosine is also negative in the third quadrant, so..

θ   ≈   360° - 131.81°   ≈   228.19°     also has a cosine of  -2/3 .

c)   (sin θ - 1)(5 cos θ + 3)   =   0

Here, we need to set each factor equal to zero and solve each for  θ  .

First factor:

sin θ - 1   =   0

sin θ   =   1         There is only one angle that has a sin of 1 .

θ   =   90°

Second factor:

5 cos θ + 3   =   0

5 cos θ   =   -3

cos θ   =   -3/5

θ   ≈   126.87°          and          θ   ≈   360° - 126.87°   ≈   233.13°

d)   tan θ   =   tan θ ( 2 + 3 sin θ )

Subtract  tan θ  from both sides.

0   =   tan θ ( 2 + 3 sin θ )  -  tan θ

Factor  tan θ  out of the terms on the right side.

0  =  (tan θ)( (2 + 3 sin θ)  -  1 )

0  =  (tan θ)( 1 + 3 sin θ)

Set each factor equal to zero and solve for  θ .

tan θ   =   0    This occurs at..

θ   =   180°     and     θ   =   360°

1 + 3 sin θ   =   0

3 sin θ   =   -1

sin θ   =   -1/3

θ   ≈   -19.47° + 360°   ≈   340.53°          and          θ   ≈   180° - -19.47°   ≈   199.47°

hectictar Dec 28, 2017
edited by hectictar  Dec 28, 2017
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Thanks

Rauhan  Dec 28, 2017
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a)  sin θ + cos θ   =   0          square both sides

sin^2 θ  +  2sin θ  cos θ   +  cos^2 θ  =    0

1  +  2sinθcosθ   =  0

2sinθcosθ    =   - 1

sin(2θ)  =  -1

let 2θ  = x

So

sin x  =  - 1        and this occurs at  x =  270°  and at  x  = 630°

So

2θ  =  270°            and       2θ  =    630°

So

θ  =  135°        and   θ  =   315°

Dec 28, 2017
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Thanks for briefly explaining this question cuz i was still a bit lost on this one even after Hectictar explained it

Rauhan  Dec 28, 2017