+0

# Maths

0
109
7
+502

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

Rauhan  Dec 27, 2017
edited by Rauhan  Dec 27, 2017

#3
+5931
+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
Sort:

#1
0

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

http://math.typeit.org/

Guest Dec 27, 2017
#2
+50
+1

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

MIRB14  Dec 27, 2017
edited by MIRB14  Dec 27, 2017
#7
+502
0

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
+5931
+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
#4
+502
+2

Thanks

Rauhan  Dec 28, 2017
#5
+81022
+2

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°

CPhill  Dec 28, 2017
#6
+502
+1

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

### 4 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details