How do you solve $\sin x+3\cos x=3?$
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.
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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