Find all real numbers in the interval [ 0 , 2pi ) that satisfy each equation. Round approximate answers to the nearest tenth.

1. sin 2x - sin x cos x = cos x

2. sin 2x cos x - cos 2x sin x = -1/2

3. cos^2(theta / 2) = sec (theta)

4. sin (2x) = 3 sin(x)

Thanks.

Guest Apr 30, 2019

#1**+2 **

1) \(sin(2x)-sin(x)cos(x)=cos(x)\)

For the sin (2x), you can use the Double Angle Formula of \(sin(2x)=2sin(x)cos(x)\).

\(2sin(x)cos(x)-sin(x)cos(x)=cos(x)\)

Then the \(2sin(x)cos(x)-sin(x)cos(x)\) simplifies to \(sin(x)cos(x)\).

\(sin(x)cos(x)=cos(x)\)

Then subtract cos(x) from both sides.

\(sin(x)cos(x)-cos(x)=0\)

You can factor the cos(x) out.

\(cos(x)(sin(x)-1)=0\)

Then, set both terms equal to 0.

\(cos(x)=0,sin(x)-1=0\)

Add 1 to both sides on the sin(x) function.

\(cos(x)=0,sin(x)=1\)

Then, take the inverse of both functions.

\(x=cos^{-1}(0),x=sin^{-1}(1)\)

The only values that satisfy those are \(x=90^o, x=270^o\) or \(x=\frac{\pi}{2},x=\frac{3\pi}{2}\).

MemeLord Apr 30, 2019