Is there any algebraic solution for the following trigonometrical equation?

The approximation $\cos \theta=1-\dfrac{(\theta) ^2}{2}$ yields the approximate equality $6\theta^2-15\theta+5π-12=0$ , from which we get the approximate value of $\theta$ as $\boxed {0.27814^c}$ or $\boxed {15.9362\degree}$

I don´t know if my answer is right, but here I go: We could calculate the interval to determine where´s the solution for the equation $15θ+12Cos(θ)=5π$ in this way:

If $15θ+12Cos(θ)=5π$ , then $θ=$ ${ Cos }^{ -1 }(\frac { 5\pi -15\theta }{ 12 } )$ , according to the inverse function of cosine of $x$ , $\frac{5π-15θ}{12}$ must be between -1 and 1: $-1\le \frac { 5\pi -15\theta }{ 12 } \le 1$ . Solving this inequality the answer must be between this interval: $\frac { 5\pi -12 }{ 15 } \le \theta \le \frac { 12+5\pi }{ 15 }$ .

If you choose these numbers like the answer you won´t get a solution of this equation and you´ll get that $\cos { (\frac { 5\pi \mp 12 }{ 15 } )=\pm 1 }$ , which is wrong. So we can find a number $k$ which we could add or subtract to the angle to calculate the answer, however, we can´t estimate the value of $k$ only with algebra because its value is between an interval. Therefore, I think we couldn´t find algebraically the solution.

We can find an approximated value with the Newton-Raphson method , with this $\theta \approx 0.27788$

Note: All the angles were written in radians