What is the general solution for solvefor x,z=cos(2x+y) ?

The general solution for solvefor x,z=cos(2x+y) is x=(arccos(z))/2+pin-y/2 ,x=-(arccos(z))/2+pin-y/2

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solvefor x,z=cos(2x+y)

solve for

x, z = cos (2 x + y)

x = \frac{arccos ( z )}{2} + πn - \frac{y}{2}, x = - \frac{arccos ( z )}{2} + πn - \frac{y}{2}

Solution steps

z = cos (2 x + y)

Switch sides

cos (2 x + y) = z

Apply trig inverse properties

2 x + y = arccos (z) + 2 πn, 2 x + y = - arccos (z) + 2 πn

Solve

2 x + y = arccos (z) + 2 πn : x = \frac{arccos ( z )}{2} + πn - \frac{y}{2}

Solve

2 x + y = - arccos (z) + 2 πn : x = - \frac{arccos ( z )}{2} + πn - \frac{y}{2}

x = \frac{arccos ( z )}{2} + πn - \frac{y}{2}, x = - \frac{arccos ( z )}{2} + πn - \frac{y}{2}

16 tan^{2} (x) - 1 = 0

sin^{2} (x) - sin (x) = 0, 0 \leq x \leq 2 π

sin (x + 3 7^{\circ}) = cos (2 x + 8)

sin (θ) = \frac{8}{17}, cos (θ) = \frac{15}{17}

tan (x) = 4746050429

What is the general solution for solvefor x,z=cos(2x+y) ?
The general solution for solvefor x,z=cos(2x+y) is x=(arccos(z))/2+pin-y/2 ,x=-(arccos(z))/2+pin-y/2