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

The general solution for cos^2(x)-cos(x)-1=0,0<= x<= 2pi is x=2.23703…,x=-2.23703…+2pi

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

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

x = 2.23703 \dots, x = - 2.23703 \dots + 2 π

Solution steps

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

Solve by substitution

cos (x) = \frac{1 + 5}{2}, cos (x) = \frac{1 - 5}{2}

cos (x) = \frac{1 + 5}{2}, 0 \leq x \leq 2 π :

No Solution

cos (x) = \frac{1 - 5}{2}, 0 \leq x \leq 2 π : x = arccos (\frac{1 - 5}{2}), x = - arccos (\frac{1 - 5}{2}) + 2 π

Combine all the solutions

x = arccos (\frac{1 - 5}{2}), x = - arccos (\frac{1 - 5}{2}) + 2 π

Show solutions in decimal form

x = 2.23703 \dots, x = - 2.23703 \dots + 2 π

2 cos (x) = csc (x)

cos (θ) = 0.3

sin^{5} (x) = 4 sin (x)

tan (x) = 2 + 3

sin (a) = \frac{11}{61}

What is the general solution for cos^2(x)-cos(x)-1=0,0<= x<= 2pi ?
The general solution for cos^2(x)-cos(x)-1=0,0<= x<= 2pi is x=2.23703…,x=-2.23703…+2pi