Trigonometric Equation using Minimum and Maximum value of Function - Practice Questions & MCQ

Trigonometric Equation using Minimum and Maximum values of Trigonometric Functions 

Sometimes, we use the maximum and minimum values of trigonometric functions to solve trigonometric equations.

While solving the equation of type $f(x)=g(x)$, and $x \in A$, we may come across a situation like If $x \in A, f(x)(L H S) \leq a$ and $g(x)(R H S) \geq a$, then the only condition under when LHS equals RHS: $f(x)=g(x)$ is when both equal a.

Illustrations 1

If $3 \sin a x+4 \cos x=7$, then the possible values of ' $a$ ' are
Here, we have given $3 \sin a x+4 \cos x=7$,
Now maximum value of LHS is $7(3+4=7)$, which occurs when $\sin \mathrm{ax}=1$ and $\cos x=1$
So LHS will equal RHS only when LHS is at its maximum value of 7 , which is possible only when $\sin \mathrm{ax}=1$ and $\cos \mathrm{x}=1$

So,

 $\begin{aligned} & \mathrm{ax}=(4 \mathrm{n}+1) \frac{\pi}{2} \text { and } \mathrm{x}=2 \mathrm{~m} \pi, \mathrm{n}, \mathrm{m} \in \mathbb{I} \\ & \frac{(4 \mathrm{n}+1) \pi}{2 \mathrm{a}}=2 \mathrm{~m} \pi \\ & \mathrm{a}=\frac{(4 \mathrm{n}+1)}{4 \mathrm{~m}}, \mathrm{n}, \mathrm{m} \in \mathbb{I}\end{aligned}$

Illustrations 2

$
\sin (x)+\cos (x)=2
$
Now the maximum value of LHS is 2 , which occurs when $\sin (x)=1$ and $\cos (x)=1$
So now we have to solve a system of simultaneous equations: $\sin (x)=1$ and $\cos (x)=1$
But we know that when $\sin (x)=1$, then $\cos (x)$ can take only one of the following values

$\sqrt{1-sin^2x}\,\, or\,\, -\sqrt{1-sin^2x}$, i.e. 0 only

So no value of $x$ exists which satisfies both the equations $\sin (x)=1$ and $\cos (x)=1$ simultaneously

So the given equation has no solution.