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

$2 \cos^{2} (x) + \cos (x) - 1 = 0$

Let $u = \cos (x)$ . Substitute $u$ for all occurrences of $\cos (x)$ .

$2 u^{2} + u - 1 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

Multiply by $1$ .

$2 u^{2} + 1 u - 1 = 0$

Rewrite $1$ as $- 1$ plus $2$

$2 u^{2} + (- 1 + 2) u - 1 = 0$

Apply the distributive property.

$2 u^{2} - 1 u + 2 u - 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 u^{2} - 1 u) + 2 u - 1 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (2 u - 1) + 1 (2 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

$(2 u - 1) (u + 1) = 0$

Replace all occurrences of $u$ with $\cos (x)$ .

$(2 \cos (x) - 1) (\cos (x) + 1) = 0$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.