Answer
Verified409.2k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths