What is the general solution for -sin(x)+cos(2x)=1 ?

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

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

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

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = 2 πn, x = π + 2 πn

Solution steps

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

Subtract

1

from both sides

- sin (x) + cos (2 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} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

Combine all the solutions

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = 2 πn, x = π + 2 πn

tan (θ) = - 3

cot (θ) = \frac{3}{3}

cos (θ) = - \frac{3}{5}

2 cos^{2} (x) - 7 cos (x) - 4 = 0

2 tan^{2} (t) = 3 sec (t)

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