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

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

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

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

x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 0, x = 2 π

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

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

Combine all the solutions

x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 0, x = 2 π

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

tan (θ) = - 24

tan^{2} (θ) + 5 tan (θ) + 2 = 0

cos (z) = - 2

sin (2 x) = sin (x), x, \in [0, 2 π]

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