The calculation changes `t = 0`

there `sin[ψ`

```
``````
eq1 = ω1
eq2 = ω2
```

and construct the linear combinations,

```
eq1n = Simplify[eq1Sin[eq1Sin[eq1Sin[eq1Sin[ψ
(* Cos[ψ
```

(If this were not possible, the equations themselves could not be solved in principle.)

Now replace `eq1, eq2`

by `eq1n, eq2n`

,

```
I1 = 2; I2 = 3; I3 = 4;
s = NDSolveValue[{I1*ω1'[{I1*ω1'[{I1*ω1'[{I1*ω1'
I2 * ω2 & # 39;
I3 * ω3 & # 39;
eq1n == 0, eq2n == 0,
ω3
ω1[0] == 2, ω2[0] == 3, ω3[0] == 4, ψ[0] == 0, φ[0] == 0, θ[0] == Pi / 6},
{ω1
plot[Evaluate@s[[1 ;; 3]], {t, 0, 120}, ImageSize -> Large]plot[Evaluate@s[[4 ;; 6]], {t, 0, 120}, ImageSize -> Large]
```

Incidentally, the original equations can be solved by slightly changing the initial state `ψ[0] == 0`

to `ψ[0] == 10 ^ -6`

,

Another approach is to use the option.

```
Method -> {"EquationSimplification" -> "Residual"}
```

Everyone gives the same answer.

```
```