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

The general solution for 2cos(x)sin(x)=1 is x= pi/4+pin

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2cos(x)sin(x)=1

2 cos (x) sin (x) = 1

x = \frac{π}{4} + πn

Solution steps

2 cos (x) sin (x) = 1

Rewrite using trig identities

sin (2 x) = 1

General solutions for

sin (2 x) = 1

2 x = \frac{π}{2} + 2 πn

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

cos (2 x + 15) = 0.3

\frac{sin ^{2} ( x )}{cos ( x )} = 16.33

- 2 sin (θ) + 2 sin^{2} (θ) = 0

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

0 = - (sin (6.9256 - x) + (sin (- x)))

What is the general solution for 2cos(x)sin(x)=1 ?
The general solution for 2cos(x)sin(x)=1 is x= pi/4+pin