+0

# Trigonometric Equations

0
158
1

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.

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 $$x=\frac{\pi}{2},x=\frac{3\pi}{2}$$.

Apr 30, 2019