What is the general solution for solvefor y,arccot(y)=ln(sin^2(3x))-e^x ?

The general solution for solvefor y,arccot(y)=ln(sin^2(3x))-e^x is y=cot(ln(sin^2(3x))-e^x)

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solvefor y,arccot(y)=ln(sin^2(3x))-e^x

solve for

y, arccot (y) = ln (sin^{2} (3 x)) - e^{x}

y = cot (ln (sin^{2} (3 x)) - e^{x})

Solution steps

arccot (y) = ln (sin^{2} (3 x)) - e^{x}

Apply trig inverse properties

arccot (y) = ln (sin^{2} (3 x)) - e^{x}

arccot (x) = a \Rightarrow x = cot (a)

y = cot (ln (sin^{2} (3 x)) - e^{x})

y = cot (ln (sin^{2} (3 x)) - e^{x})

tan^{2} (4 x) = 0

cos (\frac{x}{2}) = - 0.4

tan (x - 4 5^{\circ}) - tan (x + 4 5^{\circ}) = 4

sin (x) (2 sin^{2} (x) - 1) = 0

sin (2 x) = 10 cos (x)

What is the general solution for solvefor y,arccot(y)=ln(sin^2(3x))-e^x ?
The general solution for solvefor y,arccot(y)=ln(sin^2(3x))-e^x is y=cot(ln(sin^2(3x))-e^x)