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

The general solution for cos(2x)-sin(x)=1,0<= x<2pi is x=(7pi)/6 ,x=(11pi)/6 ,x=0,x=pi

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

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

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0, x = π

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0, x = π

sec^{2} (x) + 5 sec (x) - 14 = 0

3 cos^{2} (x) = 1

cos (θ) = - \frac{7}{8}, csc (θ)

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

tan (θ) = - 3, sin (θ) < 0

What is the general solution for cos(2x)-sin(x)=1,0<= x<2pi ?
The general solution for cos(2x)-sin(x)=1,0<= x<2pi is x=(7pi)/6 ,x=(11pi)/6 ,x=0,x=pi