Question
Find sin θ such that $3 \cos \theta + 4 \sin \theta = 4.$

Answer

$ 3 \cos \theta+4 \sin \theta=4$
$\therefore 3 \cos \theta=4(1-\sin \theta) $
Squaring both the sides, we get.
$ 9 \cos ^2 \theta=16(1-\sin \theta)^2$
$\therefore 9\left(1-\sin ^2 \theta\right)=16\left(1+\sin ^2 \theta-2 \sin \theta\right)$
$\therefore 9-9 \sin ^2 \theta=16+16 \sin ^2 \theta-32 \sin \theta$
$\therefore 25 \sin ^2 \theta-32 \sin \theta+7=0$
$\therefore 25 \sin ^2 \theta-25 \sin \theta-7 \sin \theta+7=0$
$25 \sin \theta(\sin \theta-1)-7(\sin \theta-1)=0$
$\therefore(\sin \theta-1)(25 \sin \theta-7)=0$
$\therefore \sin \theta-1=0 \text { or } 25 \sin \theta-7=0$
$\therefore \sin \theta=1 \text { or } \sin \theta=\frac{7}{25}$
$\text { Since, }-1 \leq \sin \theta \leq 1$
$\therefore \sin \theta=1 \text { or } \frac{7}{25} $
[Note: Answer given in the textbook is 1 . However, as per our calculation it is 1 or $\frac{7}{25}$.]

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