Mastering Trig Equations: Solving for x in cosx(2sinx+1) = 0

[LordofDirT]

cosx(2sinx+1) = 0 ...it looks so easy.


method 1:

distribute: 2sinxcosx + cosx = 0

double angle: sin2x + cosx = 0

sin2x = -cosx

here I'm getting stuck.

Method 2:

I try to use the zero product property

cosx = 0
2sinx + 1 = 0

x = pi/2 + 2pi(k) , 3pi/2 + 2pi(k)

x = pi/6 + 2pi(k) , 5pi/6 + 2pi(k)

Where am I going wrong?

 

[rocomath]

If I were to ask you to solve for the roots of x for this equation, how would you go about it?

$$x(x+1)=0$$

Apply the same method and it's solved! So Method 1 should be tossed out!

You have the correct answers for Method 2.

$$\cos x=0$$

$$x=\frac{\pi}{2}, \frac{3\pi}{2}=\frac{\pi}{2}+k\pi$$

 

[LordofDirT]

Double angle equation (trig)

sin2x + cosx = 0

Attempt:

2sinxcosx + cosx = 0

sinx = -cosx/2cosx

sinx = -1/2

This gives me two solutions

x = 7pi/6 , 11pi/6 in the interval [0 , 2pi)

But the book gives 4...

 

[rock.freak667]

sin2x + cosx = 0

Attempt:

2sinxcosx + cosx = 0

sinx = -cosx/2cosx

2sinxcosx+cosx=0
cosx(2sinx-1)=0


Don't divide by a trig function unless they told you that cosx$\neq$0

Now you have a product. Each one is equal to zero. Solve now.