What is the general solution for 4cos^2(x)-4cos(x)-1=0,0<= x<= 2pi ?

The general solution for 4cos^2(x)-4cos(x)-1=0,0<= x<= 2pi is x=1.77941…,x=-1.77941…+2pi

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4cos^2(x)-4cos(x)-1=0,0<= x<= 2pi

4 cos^{2} (x) - 4 cos (x) - 1 = 0, 0 \leq x \leq 2 π

x = 1.77941 \dots, x = - 1.77941 \dots + 2 π

Solution steps

4 cos^{2} (x) - 4 cos (x) - 1 = 0, 0 \leq x \leq 2 π

Solve by substitution

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

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

No Solution

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

Combine all the solutions

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

Show solutions in decimal form

x = 1.77941 \dots, x = - 1.77941 \dots + 2 π

10 = 6.7 + 13.75 tan (θ) - 4.718 (1 + tan^{2} (θ))

2 cos (3 θ) = - 3

cos (θ) = - cos (θ)

cot (x) cos (x) = cos (x)

tan^{4} (θ) - 20 tan^{2} (θ) + 64 = 0

What is the general solution for 4cos^2(x)-4cos(x)-1=0,0<= x<= 2pi ?
The general solution for 4cos^2(x)-4cos(x)-1=0,0<= x<= 2pi is x=1.77941…,x=-1.77941…+2pi