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(1) Check the validity of the given equation, e.g. \[2\sin \theta -\cos \theta =4\] can never be true for any \[\theta \] as the value \[(2\sin \theta -\cos \theta )\] can never exceeds \[\sqrt{{{2}^{2}}+{{(-1)}^{2}}}=\sqrt{5}\]. So there is no solution of this equation.
(2) Equation involving \[\sec \theta \] or \[\tan \theta \] can never have a solution of the form,\[(2n+1)\frac{\pi }{2}\].
Similarly, equations involving \[\text{cosec }\theta \] or \[\cot \theta \] can never have a solution of the form \[\theta =n\pi \]. The corresponding functions are undefined at these values of \[\theta \].
(3) If while solving an equation we have to square it, then the roots found after squaring must be checked whether they satisfy the original equation or not, e.g. let \[x=3\]. Squaring, we get \[{{x}^{2}}=9\] \[\therefore \] \[x=3\] and \[-\,3\] but \[x=-3\] does not satisfy the original equation \[x=3\].
(4) Do not cancel more...
Suppose we have to find the principal value of \[\theta \] satisfying the equation \[\sin \theta =-\frac{1}{2}\].
Since \[\sin \theta \] is negative, \[\theta \] will be in 3rd or 4th quadrant. We can approach 3rd or 4th quadrant from two directions. If we take anticlockwise direction the numerical value of the angle will be greater than \[\pi \]. If we approach it in clockwise direction the angle will be numerically less than \[\pi \]. For principal value, we have to take numerically smallest angle. So for principal value.
(1) If the angle is in 1st or 2nd quadrant we must select anticlockwise direction and if the angle is in 3rd or 4th quadrant, we must select clockwise direction.
(2) Principal value is never numerically greater than \[\pi \].
(3) more...
In \[a\cos \theta \,+b\sin \theta =c,\] put \[a=r\,\cos \alpha \] and \[b=r\,\sin \alpha \]where \[r=\sqrt{{{a}^{2}}+{{b}^{2}}}\] and \[|c|\le \sqrt{{{a}^{2}}+{{b}^{2}}}\]
Then,\[r\,(\cos \alpha \,\cos \theta +\sin \alpha \,\sin \theta )=c\]
\[\Rightarrow \,\,\cos (\theta -\alpha )=\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\cos \beta \], (say) .....(i)
\[\Rightarrow \,\,\theta -\alpha =2n\pi \pm \beta \Rightarrow \,\theta =2n\pi \pm \beta +\alpha ,\] where \[\tan \alpha =\frac{b}{a}\] is the general solution.
Alternatively, putting \[a=r\,\sin \alpha \] and \[b=r\,\cos \alpha \],
where \[r=\sqrt{{{a}^{2}}+{{b}^{2}}}\] \[\Rightarrow \,\,\sin (\theta +\alpha )=\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\sin \gamma \], (say)
\[\Rightarrow \,\,\theta +\alpha =n\pi +{{(-1)}^{n}}\gamma \]\[\Rightarrow \,\,\theta =n\pi +{{(-1)}^{n}}\gamma -\alpha ,\]
where \[\tan \alpha =\frac{a}{b}\] is the general solution.
\[(-\sqrt{{{a}^{2}}+{{b}^{2}}})\le \,a\cos \theta +b\sin \theta \le \,(\sqrt{{{a}^{2}}+{{b}^{2}}})\]
The general solution of \[a\cos x+b\sin x=c\] is
\[x=2n\pi +{{\tan }^{-1}}\left( \frac{b}{a} \right)\pm {{\cos }^{-1}}\left( \frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)\].
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