Problem Study : Trigonometric Equations

 

Solve the equation

(i) $ \displaystyle 3\sin x-5\cos x=0$ for $ \displaystyle 0{}^\circ < x < 360{}^\circ$.

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Let $ \displaystyle R\cos \theta =3$ and $ \displaystyle R\sin \theta =5$.

$ \displaystyle \therefore R=\sqrt{{{{3}^{2}}+{{5}^{2}}}}=\sqrt{{34}}$

$ \displaystyle \ \ \ \tan \theta =\frac{5}{3}=1.6667$

$ \displaystyle \therefore \theta =59{}^\circ {2}'$

$ \displaystyle \therefore 3\sin x-5\cos x=\sqrt{{34}}\sin (x-59{}^\circ {2}')$

$ \displaystyle \therefore \sqrt{{34}}\sin (x-59{}^\circ {2}')=0$

$ \displaystyle \therefore \sin ((x-59{}^\circ {2}')=0$

$ \displaystyle \therefore x-59{}^\circ {2}'=0{}^\circ $ (or) $ \displaystyle x-59{}^\circ {2}'=180{}^\circ $

$ \displaystyle \therefore x=59{}^\circ {2}'$ (or) $ \displaystyle x=239{}^\circ {2}'$


(ii) $ \displaystyle 5{{\sin }^{2}}y+9\cos y-3=0\ \text{for }$ $ \displaystyle 0{}^\circ < y < 360{}^\circ$.

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$ \displaystyle 5(1-{{\cos }^{2}}y)+9\cos y-3=0$

$ \displaystyle 5-5{{\cos }^{2}}y+9\cos y-3=0$

$ \displaystyle 5{{\cos }^{2}}y-9\cos y-2=0$

$ \displaystyle (5\cos y+1)(\cos y-2)=0$

$ \displaystyle \therefore \cos y=-\frac{1}{5}\ $ or $ \displaystyle \cos y=2$

Since $\displaystyle -1\le \cos y\le 1$, $ \displaystyle \cos y=2$ is impossible.

$ \displaystyle \therefore \cos y=-\frac{1}{5}=-0.2$

$ \displaystyle \therefore \text{basic acute angle = }78{}^\circ 2{8}'$

Since $ \displaystyle \cos y < 0$, $ \displaystyle y$ lies in the second or third quadrant.

$ \displaystyle \therefore y=180{}^\circ -78{}^\circ 2{8}'\ \text{or }y=180{}^\circ +78{}^\circ 2{8}'$

$ \displaystyle \therefore y=101{}^\circ 3{2}'\ \text{or }y=258{}^\circ 2{8}'$


(iii) $ \displaystyle 6{{\sin }^{2}}x=5+\cos x\ \text{for }0{}^\circ < x < 180{}^\circ .$

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$\displaystyle 6{{\sin }^{2}}x=5+\cos x$

$ \displaystyle 6(1-{{\cos }^{2}}x)=5+\cos x$

$ \displaystyle 6-6{{\cos }^{2}}x=5+\cos x$

$ \displaystyle 6{{\cos }^{2}}x+\cos x-1=0$

$ \displaystyle (3\cos x-1)(2\cos x+1)=0$

$ \displaystyle \cos x=\frac{1}{3}\ \ \text{or}\ \cos x=-\frac{1}{2}$

$ \displaystyle \therefore x=70{}^\circ 3{1}'\ \text{or}\ x=120{}^\circ $