+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) \(tan\theta = tan\theta(2+3 sin \theta )\)
+9466 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°
You can use this online Math keyboard to write the Math symbols you want:
http://math.typeit.org/
+9466 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°
