What is the general solution for cos(2x+5)=sin(2x) ?

The general solution for cos(2x+5)=sin(2x) is x=pin+pi/8-5/4 ,x=(5pi)/8-5/4+pin

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cos(2x+5)=sin(2x)

cos (2 x + 5) = sin (2 x)

x = πn + \frac{π}{8} - \frac{5}{4}, x = \frac{5 π}{8} - \frac{5}{4} + πn

Solution steps

cos (2 x + 5) = sin (2 x)

Subtract

sin (2 x)

from both sides

cos (2 x + 5) - sin (2 x) = 0

Rewrite using trig identities

- 2 sin (\frac{π + 10}{4}) sin (\frac{8 x - π + 10}{4}) = 0

Divide both sides by

- 2 sin (\frac{π + 10}{4})

sin (\frac{8 x - π + 10}{4}) = 0

General solutions for

sin (\frac{8 x - π + 10}{4}) = 0

\frac{8 x - π + 10}{4} = 0 + 2 πn, \frac{8 x - π + 10}{4} = π + 2 πn

Solve

\frac{8 x - π + 10}{4} = 0 + 2 πn : x = πn + \frac{π}{8} - \frac{5}{4}

Solve

\frac{8 x - π + 10}{4} = π + 2 πn : x = \frac{5 π}{8} - \frac{5}{4} + πn

x = πn + \frac{π}{8} - \frac{5}{4}, x = \frac{5 π}{8} - \frac{5}{4} + πn

sin (2 θ) + 3 cos (θ) = cos (θ)

1 + cos (θ) = 1 + sin (θ)

500 = 538 cos (x - 1 0^{\circ})

0 = sec^{2} (x) e^{t a n (x)} (2 tan (x) + sec^{2} (x))

tan (x) = 3, - 18 0^{\circ} < x < 18 0^{\circ}

What is the general solution for cos(2x+5)=sin(2x) ?
The general solution for cos(2x+5)=sin(2x) is x=pin+pi/8-5/4 ,x=(5pi)/8-5/4+pin