1 - 1/2 sin(2theta) = ?

1-\frac12sin(2\theta)

1. treat sin(2\theta) as a single term and solve for it normally. You'll end up with:

sin(2\theta)=2

2. However, this equation had no solutions since the sine curve never cuts the line y = 2. This ends your question.

Or, if that was: 1-\frac{1}{2sin(2\theta)}

1. Same procedure but end up with

sin(2\theta)=\frac{1}{2}

2. Find the key angle (here denoted by alpha);

\alpha=sin^{-1}(\frac{1}{2})

\alpha = 30

3. Determine the quadrants in which the angle is found, here it is in the first and second quadrant, where sine is both positive.

4. Equate 2\theta to your obtained angles, you'll have

2\theta = 30, 150

5. Finally solve for theta, you'll have 15 and 75 in degrees.

No see the answer has to be in sin and cos



1-\frac12sin(2\theta)

1. treat sin(2\theta) as a single term and solve for it normally. You'll end up with:

sin(2\theta)=2

2. However, this equation had no solutions since the sine curve never cuts the line y = 2. This ends your question.

Or, if that was: 1-\frac{1}{2sin(2\theta)}

1. Same procedure but end up with

sin(2\theta)=\frac{1}{2}

2. Find the key angle (here denoted by alpha);

\alpha=sin^{-1}(\frac{1}{2})

\alpha = 30

3. Determine the quadrants in which the angle is found, here it is in the first and second quadrant, where sine is both positive.

4. Equate 2\theta to your obtained angles, you'll have

2\theta = 30, 150

5. Finally solve for theta, you'll have 15 and 75 in degrees.

1 - \frac 12 sin(2 \theta) = ?

sin (2 \theta) = 2 (sin \theta) (cos \theta)

so

1 - \frac 12 sin(2 \theta) = 1-sin(\theta)cos(\theta)

Oh, sorry, I couldn't possibly know at what level of Mathematics you are... Perito gave the answer anyway...