Can I solve the equation like a quadratic?

2cos^2x + cos x-1=0

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Certainly, you can solve the given equation in this way.

First, you have to substitute cos x by another variable. We'll change the variable and we'll re-write the equation.

We'll put cos x = t.

We'll raise to square both sides:

(cos x)^2 = t^2

We'll re-write the equation:

2t^2 + t - 1 = 0

Now, we'll apply the quadratic formula:

t1 = [-1 + sqrt(1 + 8)]/4

t1 = (-1+3)/4

t1 = 2/4

t1 = 1/2

t2 = (-1-3)/4

t2 = -1

Now, we'll find x from:

cos x = t1

cos x = 1/2

x = arccos (1/2) + 2kpi

x = pi/3 + 2kpi

cos x = t2

cos x = -1

x = arccos(-1) + 2kpi

x = pi + 2kpi

**The solutions of the equation are: {pi/3 + 2kpi}U{pi + 2kpi}.**

To solve for 2cos^2x+cosx-1 = 0

We rewite the equation as:

2cos^2x + 2cosx -cosx -1 = 0.

We group th left as:

(2cos^2x+2cosx) - (cosx+1) = 0

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

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

Equate the factors to zero:

cosx+1 = 0

2cosx-1 = 0

cosx+1 = 0 gives cosx = 1, or x = pi/2 , 2npi+pi/2 0r x = 2npi-pi/2.

2cosx-1 = 0 gives cosx = 1/2. x = 2npi + pi/3 or x = 2npi-pi/3.

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