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

The general solution for cos(2x)-cos(x)=-1,0<= x<2pi is x= pi/3 ,x=(5pi)/3 ,x= pi/2 ,x=(3pi)/2

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

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

x = \frac{π}{3}, x = \frac{5 π}{3}, x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

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

Subtract

- 1

from both sides

cos (2 x) - cos (x) + 1 = 0

Rewrite using trig identities

- cos (x) + 2 cos^{2} (x) = 0

Solve by substitution

cos (x) = \frac{1}{2}, cos (x) = 0

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

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

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}, x = \frac{π}{2}, x = \frac{3 π}{2}

sin (x) = - \frac{1}{10}

sec^{2} (θ) + sec (θ) = 2, 2 π \leq θ \leq 3 π

6 sec^{2} (x) = 0

3 - 2 cos (x) = 2

tan (θ) = - \frac{8}{5}

What is the general solution for cos(2x)-cos(x)=-1,0<= x<2pi ?
The general solution for cos(2x)-cos(x)=-1,0<= x<2pi is x= pi/3 ,x=(5pi)/3 ,x= pi/2 ,x=(3pi)/2