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

The general solution for sin^2(x)-cos^2(x)+sin(x)=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6 ,x=(3pi)/2

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

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

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

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

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

cos (θ) = - cos (2 θ)

3 (sin (x) - cos (x)) = 2 (sin (x) + cos (x))

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

θ, ma = t sin (θ) - m g

x, sin (x) - cos (y) - 2 = 0

What is the general solution for sin^2(x)-cos^2(x)+sin(x)=0,0<= x<= 2pi ?
The general solution for sin^2(x)-cos^2(x)+sin(x)=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6 ,x=(3pi)/2