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

Nyasia Flowers
2022-09-11
Answered

Solve the Equation

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

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

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Aldo Harrington

Answered 2022-09-12
Author has **12** answers

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

$$\mathrm{cos}x=t;-1\le t\le 1$$

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

D=1+8=9

$${x}_{1}=\frac{-1+3}{4}=\frac{1}{2}$$

$${x}_{2}=\frac{-1-3}{4}=-1$$

$$\mathrm{cos}x=\frac{1}{2}$$

$$x=+-\mathrm{arccos}\frac{1}{2}+2\pi \ast n$$

$$x=+-\frac{\pi}{3}+2\pi \ast n$$

$$x=+-\frac{\pi}{3}+2\pi \ast n$$

$$\mathrm{cos}x=-1$$

$$x=+-(\pi -\mathrm{arccos}1)+2\pi \ast k$$

$$x=+-\pi +2\pi \ast k$$

n and k $$\in $$ Z.

$$\mathrm{cos}x=t;-1\le t\le 1$$

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

D=1+8=9

$${x}_{1}=\frac{-1+3}{4}=\frac{1}{2}$$

$${x}_{2}=\frac{-1-3}{4}=-1$$

$$\mathrm{cos}x=\frac{1}{2}$$

$$x=+-\mathrm{arccos}\frac{1}{2}+2\pi \ast n$$

$$x=+-\frac{\pi}{3}+2\pi \ast n$$

$$x=+-\frac{\pi}{3}+2\pi \ast n$$

$$\mathrm{cos}x=-1$$

$$x=+-(\pi -\mathrm{arccos}1)+2\pi \ast k$$

$$x=+-\pi +2\pi \ast k$$

n and k $$\in $$ Z.

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