Can you please review this to see if I did this right?

2cos^2x+3cosx=0

sq. root of (cos^2x)+(cosx)=sq. root of (-3/2)

2cos=(sq. root of -3 times sq. root of 2)/2)) times 2/1

cos=(2 times sq. root of -3 times sq. root of two)/2))

To review your work, let's solve the equation step by step.

Starting with the equation: 2cos^2x + 3cosx = 0

Step 1: Factor out cos(x):
cosx(2cosx + 3) = 0

Now, one of the factors must be zero, so either cosx = 0 or 2cosx + 3 = 0.

Step 2: Solve for cosx = 0:
cosx = 0

Step 3: Solve for 2cosx + 3 = 0:
2cosx + 3 = 0
2cosx = -3
cosx = -3/2

Now, let's check your steps:

You square rooted both sides of the equation, which is not necessary and can introduce extraneous solutions. So, squaring both sides of the equation is not recommended.

The expression you wrote after the square root symbol is incorrect. Instead, it should be the equation itself: cos^2x + cosx = -3/2.

The next step where you multiplied both sides of the equation by 2/1 is not correct either. You cannot isolate the cos term like that. Instead, you should follow the steps mentioned above.

To summarize, the correct values of cosx that satisfy the equation 2cos^2x + 3cosx = 0 are cosx = 0 and cosx = -3/2.