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

The general solution for 4cos^2(x)+2cos(x)=2,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=pi

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

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

x = \frac{π}{3}, x = \frac{5 π}{3}, x = π

Solution steps

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

Solve by substitution

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

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

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

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}, x = π

arctan (x) + arctan (2 x) = \frac{π}{4}

1 + cos (θ) = \frac{2 + 3}{2}

1 + 3 sin (x) = 2

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

sinh (x) = \frac{2}{2}

What is the general solution for 4cos^2(x)+2cos(x)=2,0<= x<= 2pi ?
The general solution for 4cos^2(x)+2cos(x)=2,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=pi