Solve for theta over the interval 0 to 360

Question

5 sin theta + 5rad2=2rad2 

Answer 1

I think your equation is: 5sin(θ) + 5√2 = 2√2

 

Solve like this:

 

5sin(θ) + 5√2 = 2√2

5sin(θ) = 2√2 - 5√2

5sin(θ) = -3√2

sin(θ) = (-3√2)/5

θ = sin^-1((-3√2)/5)

θ ≈ -58.05° = 301.95°

 

This angle is between 270° and 360°, so it's located in the Fourth Quadrant. Sine is also negative in the Third Quadrant, so we'll need to find this angle's reflection across the y-axis into the Third Quadrant.

 

θ is 58.05° less than 360°, so its reflection is 180° + 58.05° = 238.05°. Draw this on the unit circle to get a better idea of why that works.

 

So, your two answers are 238.05° and 301.95°.