What is the general solution for 2cos^2(x)+2cos(x)-1=0,0<= x<= 2pi ?

The general solution for 2cos^2(x)+2cos(x)-1=0,0<= x<= 2pi is x=1.19606…,x=2pi-1.19606…

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2cos^2(x)+2cos(x)-1=0,0<= x<= 2pi

2 cos^{2} (x) + 2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

x = 1.19606 \dots, x = 2 π - 1.19606 \dots

Solution steps

2 cos^{2} (x) + 2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

Solve by substitution

cos (x) = \frac{- 1 + 3}{2}, cos (x) = - \frac{1 + 3}{2}

cos (x) = \frac{- 1 + 3}{2}, 0 \leq x \leq 2 π : x = arccos (\frac{3 - 1}{2}), x = 2 π - arccos (\frac{3 - 1}{2})

cos (x) = - \frac{1 + 3}{2}, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = arccos (\frac{3 - 1}{2}), x = 2 π - arccos (\frac{3 - 1}{2})

Show solutions in decimal form

x = 1.19606 \dots, x = 2 π - 1.19606 \dots

sec (x) = \frac{7}{5}

cos (5 θ) = sin (3 θ)

(2 sin (2 x) + 2 cos (2 x))^{2} = 4

sin (x) = \frac{7}{3}

0 = sin (4 x)

What is the general solution for 2cos^2(x)+2cos(x)-1=0,0<= x<= 2pi ?
The general solution for 2cos^2(x)+2cos(x)-1=0,0<= x<= 2pi is x=1.19606…,x=2pi-1.19606…