if 3 sin theta+4 cos theta=5 then find the value of sin theta

We have :

$\Rightarrow 3\sin \theta +4\cos \theta =5$

Subtract $3\sin \theta$ from both sides $\to$

$\Rightarrow 4\cos \theta =5-3\sin \theta$

Square both sides $\to$

$\Rightarrow 16{\cos }^2\theta =25+9{\sin }^2\theta -2\left(5\right)\left(3\sin \theta \right)$

Now substitute ${\cos }^2\theta =1-{\sin }^2\theta$ at L.H.S $\to$

$\Rightarrow 16(1-{\sin }^2\theta )=25+9{\sin }^2\theta -30\sin \theta$

Rearrange the terms to get a quadratic in $\sin \theta \to$

$\Rightarrow 9{\sin }^2\theta +16{\sin }^2\theta -30\sin \theta +25-16=0$

$\Rightarrow 25{\sin }^2\theta -30\sin \theta +9=0$

Now if you look , the above quadratic equation is a perfect square using the identity $a^2+b^2-2ab={(a-b)}^2$ we write it as $\to$

$\Rightarrow {{{\left(5\sin \theta \right)}^{}}^{}}^{}\underset{a^2}{\underset{}{{}^{}}}-{{2\cdot }^{}}^{}\underset{-2ab}{\underset{}{\left(5\sin \theta \right)\cdot \left(3\right)}}+\underset{b^2}{{{\left(3\right)}^{}}^{}}=0$

$\Rightarrow {(5\sin \theta -3)}^2=0$

$\Rightarrow 5\sin \theta -3=0$

$\Rightarrow \sin \theta =\frac{3}{5}$ .