More trig help

[Jordannn1990]

Hello, I am back again for help on a few more problems

1. Solve for θ if 0° ≤ θ < 360°. sin(theta/2) + cos(theta)=0

2. Solve the equation for x if 0 ≤ x < 2π. 2 sinx + cotx- cscx=0

3. Find all degree solutions in the interval 0° ≤ θ < 360°. If rounding is necessary, round to the nearest tenth of a degree.

6sin^2(theta) - 5cos(2theta)=0


Can anyone help me with these?

 

[mmm4444bot]

Did you miss the main points in our posting guidelines? It's best to start a new thread for each exercise.

What have you tried? Where are you stuck?

On the first exercise, did you apply the 1/2-angle identity, to rewrite sin(theta/2)?

On the second exercise, did you rewrite everything in terms of sine and cosine?

Looks like you could consider a couple of identities in the third exercise...

 

[MarkFL]

Alternate approaches:

1.) Use the double-angle identity for cosine to get a quadratic in \(\displaystyle \sin\left(\frac{\theta}{2} \right)\).

2.) Multiply the equation by \(\displaystyle \sin(x)\) and then apply a Pythagorean identity to get a quadratic in \(\displaystyle \cos(x)\).

3.) Use the double-angle identity for cosine to get a quadratic in \(\displaystyle \sin(\theta)\).

Post your work if you get stuck, and we will be glad to help.

 

[Jordannn1990]

2. Solve the equation for x if 0 ≤ x < 2π. 2 sinx + cotx- cscx=0

2sinx + cos/sin - 1/sin=0 Is this a start?



3. Find all degree solutions in the interval 0° ≤ θ < 360°. If rounding is necessary, round to the nearest tenth of a degree.

6sin^2(theta) - 5cos(2theta)=0

I got it down to 16sin^2-5=0



1. Solve for θ if 0° ≤ θ < 360°. sin(theta/2) + cos(theta)=0 I solved for this one and got 180degrees

 

[Deleted member 4993]

2. Solve the equation for x if 0 ≤ x < 2π. 2 sinx + cotx- cscx=0

2sin(x) + cos(x) /sin(x) - 1/sin(x) = 0 Is this a start?

Yes... now assuming \(\displaystyle sin(x) \ \ne \ 0 \)

multiply both sides by sin(x)


3. Find all degree solutions in the interval 0° ≤ θ < 360°. If rounding is necessary, round to the nearest tenth of a degree.

6sin^2(theta) - 5cos(2theta)=0

I got it down to 16sin^2(x) -5=0

Now

16 sin²(x) = 5

and continue...



1. Solve for θ if 0° ≤ θ < 360°. sin(theta/2) + cos(theta)=0 I solved for this one and got 180degrees..........Correct

.

 

[Jordannn1990]

.

2. I got 2pi/3 and 4pi/3


3. I got 34, 146 , and 326. What is the last angle?

 

[Deleted member 4993]

2. I got 2pi/3 and 4pi/3


3. I got 34, 146 , and 326. What is the last angle?

use

sin(Θ) = sin(π-Θ) and

sin(-Θ) = sin(2π-Θ)