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

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

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

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

x = π

Solution steps

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

Solve by substitution

cos (x) = - 1

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

Combine all the solutions

x = π

x, sin (2 x) = \frac{1}{2}

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

2 sin (θ) cos (θ) - cos (θ) = 0

sin (2 x) cot (x) = 0

3 sec (θ) - 23 = 0

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