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

The general solution for sin(x)-cos(2x)-1=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin

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

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

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = \frac{- 1 + 17}{4}, sin (x) = \frac{- 1 - 17}{4}

sin (x) = \frac{- 1 + 17}{4} : x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{- 1 - 17}{4} :

No Solution

Combine all the solutions

x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

Show solutions in decimal form

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

sin^{2} (x) = 3

x, c^{2} = a^{2} + b^{2} - 2 ab \cdot cos (x)

sin (- \frac{π}{8} c) = - \frac{14}{14}

\frac{25}{sin ( x )} = \frac{18}{sin ( 3 6 ^{\circ} )}

7 tan (3 x) - 21 tan (x) = 0

What is the general solution for sin(x)-cos(2x)-1=0 ?
The general solution for sin(x)-cos(2x)-1=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin