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Question: How do you solve \(\sin x+3\cos x=3?\)...

How do you solve sinx+3cosx=3?\sin x+3\cos x=3?

Explanation

Solution

Here, a trigonometric equation is given which we have to simplify.
Here, we are using a basic formula for simplifying the equation i.e. (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}
We also have to use a trigonometric functions of sin2x+cos2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1
And after solving the equation, we will get the value of xx in the form of angles and convert it into radians. We can check the value of xx by substituting the value of xx in the given equation.

Complete step by step solution:
In this numerical a trigonometric equation is given which is as follows:
sinx+3cosx=3...(i)\sin x+3\cos x=3...(i)
Squaring the equation (i)(i) for getting some identities to solve further.
(sinx+3cosx)2=(3)2{{\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 (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}
sin2x+2×3sinxcosx+(3)2cos2x=9{{\sin }^{2}}x+2\times 3\sin x\cos x+{{\left( 3 \right)}^{2}}{{\cos }^{2}}x=9
sin2x+6sinxcosx+9cos2x=9\Rightarrow {{\sin }^{2}}x+6\sin x\cos x+9{{\cos }^{2}}x=9
Now transpose 9cos2x9{{\cos }^{2}}x to the right side sin2x+6sincosx=99cos2x{{\sin }^{2}}x+6\sin \cos x=9-9{{\cos }^{2}}x
sin2x+6sinxcosx=9(1cos2x){{\sin }^{2}}x+6\sin x\cos x=9\left( 1-{{\cos }^{2}}x \right)
As, we know that,
sin2x+cos2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1
sin2x=1cos2x\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x
sin2x+6sinxcosx=9sin2x\Rightarrow {{\sin }^{2}}x+6\sin x\cos x=9{{\sin }^{2}}x
Subtract sin2x{{\sin }^{2}}x on the both sides from the above equation. We get,
sin2x+6sinxcosxsin2x=9sin2xsin2x{{\sin }^{2}}x+6\sin x\cos x-{{\sin }^{2}}x=9{{\sin }^{2}}x-{{\sin }^{2}}x
6sinxcosx=8sin2x6\sin x\cos x=8{{\sin }^{2}}x
Now separate the like terms,
sinxcosxsin2x=86\dfrac{\sin x\cos x}{{{\sin }^{2}}x}=\dfrac{8}{6}
cosxsinx=86\Rightarrow \dfrac{\cos x}{\sin x}=\dfrac{8}{6}
cotx=43\Rightarrow \cot x=\dfrac{4}{3}
We know that,
cotx=1tanx=tan1x\cot x=\dfrac{1}{\tan x}={{\tan }^{-1}}x
x=tan1(34)...(ii)\Rightarrow x={{\tan }^{-1}}\left( \dfrac{3}{4} \right)...(ii)
x=36.80\Rightarrow x=36.80
x=0.6435x=0.6435 radians

Therefore the value of xx for the given trigonometric equation is 0.64350.6435 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 44 basic equations or functions:
sinx=a,cosx=a\sin x=a,\cos x=a
tanx=a,cotx=a\tan x=a,\cot x=a
Using above basic functions of trigonometry other simplified equations can be make.

Note: In equation (ii)(ii)
x=tan1(34)x={{\tan }^{-1}}\left( \dfrac{3}{4} \right)
But the original equation is,
cotx=43\cot x=\dfrac{4}{3}
1tanx=cotx\Rightarrow\dfrac{1}{\tan x}=\cot x
tan1x=cotx{{\tan }^{-1}}x=\cot x
Now, tan1(34){{\tan }^{-1}}\left( \dfrac{3}{4} \right)
x=tan1(34)x={{\tan }^{-1}}\left( \dfrac{3}{4} \right)
=0.6435=0.6435 radians
We can solve this questions in other way too which is as follows:
The given equation is,
sinx+3cosx=3...(i)\sin x+3\cos x=3...(i)
Now, put tana=3\tan a=3
a=tan1(3)a={{\tan }^{-1}}\left( 3 \right)
a=71.56a=71.56
And cosa=cos(71.56)=0.32\cos a=\cos \left( 71.56 \right)=0.32
Now, put sinacosa\dfrac{\sin a}{\cos a} at the place of 33 in left side only of the equation (iii)(iii)
(sinacosa)cosx=3...(ii)\left( \dfrac{\sin a}{\cos a} \right)\cos x=3...(ii)
Multiply cosa\cos a at both sides of equation (ii)(ii)
sinxcosa+(sinacosa)cosacosx=3cosa\sin x\cos a+\left( \dfrac{\sin a}{\cos a} \right)\cos a\cos x=3\cos a
sinxcosa+sinacosx=3cosa\Rightarrow\sin x\cos a+\sin a\cos x=3\cos a
sin(x+a)=3cosa...(iii)\Rightarrow\sin \left( x+a \right)=3\cos a...(iii)
But as we already calculated above
a=71.56a=71.56
cosa=0.32\cos a=0.32
The equation (iii)(iii) becomes
sin(x+71.56)=3(0.32)\sin \left( x+71.56 \right)=3\left( 0.32 \right)
sin(x+71.56)=0.96\Rightarrow\sin \left( x+71.56 \right)=0.96
The angle should be of 180180{}^\circ
x+71.56=0x+71.56{}^\circ =0
x+71.56=18071.56\Rightarrow x+71.56{}^\circ =180{}^\circ -71.56{}^\circ
=108.44\Rightarrow=108{}^\circ .44{}^\circ
x=108.4471.56x=108{}^\circ .44{}^\circ -71.56{}^\circ x=36.88x=36.88{}^\circ
We can also verify the answer.
Put value of x=36.86x=36.86{}^\circ or x=0.6435x=0.6435 in equation (i)(i)
sinx+3cosx=3\sin x+3\cos x=3
sin(36.86)+3cos(36.86)=3\Rightarrow\sin \left( 36.86 \right)+3\cos \left( 36.86 \right)=3
0.6+3×0.8=3\Rightarrow0.6+3\times 0.8=3
0.6+2.4=3\Rightarrow0.6+2.4=3
3=33=3
Or sinx+3cosx=3\sin x+3\cos x=3
sin(0.6435)+3cos(0.6435)=3\sin \left( 0.6435 \right)+3\cos \left( 0.6435 \right)=3
0.012+3×2.9997=3\Rightarrow0.012+3\times 2.9997=3
3.0109=3\Rightarrow3.0109=3
333\simeq 3
From above it is clear that the value of x=36.86x=36.86{}^\circ or x=0.6435x=0.6435 radians is correct.