Answer

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**Hint:**Here, a trigonometric equation is given which we have to simplify.

Here, we are using a basic formula for simplifying the equation i.e. ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

We also have to use a trigonometric functions of ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

And after solving the equation, we will get the value of $x$ in the form of angles and convert it into radians. We can check the value of $x$ by substituting the value of $x$ in the given equation.

**Complete step by step solution:**

In this numerical a trigonometric equation is given which is as follows:

$\sin x+3\cos x=3...(i)$

Squaring the equation $(i)$ for getting some identities to solve further.

${{\left( \sin x+3\cos x \right)}^{2}}={{\left( 3 \right)}^{2}}$

Now, simplify the above equation. The left hand side is in the form of ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

${{\sin }^{2}}x+2\times 3\sin x\cos x+{{\left( 3 \right)}^{2}}{{\cos }^{2}}x=9$

$\Rightarrow {{\sin }^{2}}x+6\sin x\cos x+9{{\cos }^{2}}x=9$

Now transpose $9{{\cos }^{2}}x$ to the right side ${{\sin }^{2}}x+6\sin \cos x=9-9{{\cos }^{2}}x$

${{\sin }^{2}}x+6\sin x\cos x=9\left( 1-{{\cos }^{2}}x \right)$

As, we know that,

${{\sin }^{2}}x+{{\cos }^{2}}x=1$

$\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x$

$\Rightarrow {{\sin }^{2}}x+6\sin x\cos x=9{{\sin }^{2}}x$

Subtract ${{\sin }^{2}}x$ on the both sides from the above equation. We get,

${{\sin }^{2}}x+6\sin x\cos x-{{\sin }^{2}}x=9{{\sin }^{2}}x-{{\sin }^{2}}x$

$6\sin x\cos x=8{{\sin }^{2}}x$

Now separate the like terms,

$\dfrac{\sin x\cos x}{{{\sin }^{2}}x}=\dfrac{8}{6}$

$\Rightarrow \dfrac{\cos x}{\sin x}=\dfrac{8}{6}$

$\Rightarrow \cot x=\dfrac{4}{3}$

We know that,

$\cot x=\dfrac{1}{\tan x}={{\tan }^{-1}}x$

$\Rightarrow x={{\tan }^{-1}}\left( \dfrac{3}{4} \right)...(ii)$

$\Rightarrow x=36.80$

$x=0.6435$ radians

**Therefore the value of $x$ for the given trigonometric equation is $0.6435$ radians.**

**Additional Information:**

For solving a trigonometric equation we have to transform that trigonometric equation into one or more than one basic trigonometric function or equations.

It means that solving a trigonometric equation is nothing but solving one or more basic trigonometric equations or functions.

Trigonometric equations or functions.

Trigonometric equations has $4$ basic equations or functions:

$\sin x=a,\cos x=a$

$\tan x=a,\cot x=a$

Using above basic functions of trigonometry other simplified equations can be make.

**Note:**In equation $(ii)$

$x={{\tan }^{-1}}\left( \dfrac{3}{4} \right)$

But the original equation is,

$\cot x=\dfrac{4}{3}$

$\Rightarrow\dfrac{1}{\tan x}=\cot x$

${{\tan }^{-1}}x=\cot x$

Now, ${{\tan }^{-1}}\left( \dfrac{3}{4} \right)$

$x={{\tan }^{-1}}\left( \dfrac{3}{4} \right)$

$=0.6435$ radians

We can solve this questions in other way too which is as follows:

The given equation is,

$\sin x+3\cos x=3...(i)$

Now, put $\tan a=3$

$a={{\tan }^{-1}}\left( 3 \right)$

$a=71.56$

And $\cos a=\cos \left( 71.56 \right)=0.32$

Now, put $\dfrac{\sin a}{\cos a}$ at the place of $3$ in left side only of the equation $(iii)$

$\left( \dfrac{\sin a}{\cos a} \right)\cos x=3...(ii)$

Multiply $\cos a$ at both sides of equation $(ii)$

$\sin x\cos a+\left( \dfrac{\sin a}{\cos a} \right)\cos a\cos x=3\cos a$

$\Rightarrow\sin x\cos a+\sin a\cos x=3\cos a$

$\Rightarrow\sin \left( x+a \right)=3\cos a...(iii)$

But as we already calculated above

$a=71.56$

$\cos a=0.32$

The equation $(iii)$ becomes

$\sin \left( x+71.56 \right)=3\left( 0.32 \right)$

$\Rightarrow\sin \left( x+71.56 \right)=0.96$

The angle should be of $180{}^\circ $

$x+71.56{}^\circ =0$

$\Rightarrow x+71.56{}^\circ =180{}^\circ -71.56{}^\circ $

$\Rightarrow=108{}^\circ .44{}^\circ $

$x=108{}^\circ .44{}^\circ -71.56{}^\circ $$x=36.88{}^\circ $

We can also verify the answer.

Put value of $x=36.86{}^\circ $ or $x=0.6435$ in equation $(i)$

$\sin x+3\cos x=3$

$\Rightarrow\sin \left( 36.86 \right)+3\cos \left( 36.86 \right)=3$

$\Rightarrow0.6+3\times 0.8=3$

$\Rightarrow0.6+2.4=3$

$3=3$

Or $\sin x+3\cos x=3$

$\sin \left( 0.6435 \right)+3\cos \left( 0.6435 \right)=3$

$\Rightarrow0.012+3\times 2.9997=3$

$\Rightarrow3.0109=3$

$3\simeq 3$

From above it is clear that the value of $x=36.86{}^\circ $ or $x=0.6435$ radians is correct.

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