
The equation \[3\cos x + 4\sin x = 6\] has
A. Finite solution
B. Infinite solution
C. One solution
D. No solution
Answer
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Hints
In the following question, a sine and cosine trigonometric expression is supplied, and we are asked to determine its minimal value. We need to understand how to calculate the range of the trigonometric expression of type (\[asinx + bcosx\]) in order to solve the problem. The range of the formula \[3cosx + 4sinx\]is first calculated.
Formula used:
\[a\cos x + b\sin x = c\]
Complete step-by-step solution
We have given the trigonometric equation,
\[3\cos x + 4\sin x = 6\]
Thus, the given equation is of the form
\[a\cos x + b\sin x = c\]
From the equation, we get
\[a = 3,b = 4\]and \[{\rm{c}} = 6\].
Let: \[a = 3 = r\cos \alpha \]and \[b = 4 = r\sin \alpha \]
Now, the expression using the values can be written as
\[\tan \alpha = \frac{b}{a} = \frac{4}{3}\]
Solve for\[\alpha \], by moving tan to the other side to calculate the value of\[\alpha \]:
\[ \Rightarrow \alpha = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)\]
Also, we know that
\[r = \sqrt {{a^2} + {b^2}} \]
Substitute the value of \[a\]and\[b\]:
\[ = \sqrt {9 + 16} \]
Add the terms inside the root to simplify the expression:
\[ = \sqrt {25} \]
Take square root of the obtained expression to solve the value for\[r\]:
\[r = 5\]
On putting the values of \[a = 3 = r\cos \alpha \] and \[b = 4 = r\sin \alpha \] in equation (i), we get:
\[r\cos \alpha \cos \theta + \sin \alpha \sin \theta = 6\]
This can be written as
\[ \Rightarrow r\cos (\theta - \alpha ) = 6\]
As both the roots are imaginary there exists no value of \[x\] satisfying this given equation.
Therefore, the equation \[3\cos x + 4\sin x = 6\] has no sol ution.
Hence, the option D is correct.
Note
Trigonometry and inequality ideas must be thoroughly understood to answer such queries. To solve the presented problem, one must be familiar with the methods for computing the range of trigonometric expressions of the form (\[asinx + bcosx\]). To solve such issues, we also need to understand how to calculate the range of a composite complex number.
In the following question, a sine and cosine trigonometric expression is supplied, and we are asked to determine its minimal value. We need to understand how to calculate the range of the trigonometric expression of type (\[asinx + bcosx\]) in order to solve the problem. The range of the formula \[3cosx + 4sinx\]is first calculated.
Formula used:
\[a\cos x + b\sin x = c\]
Complete step-by-step solution
We have given the trigonometric equation,
\[3\cos x + 4\sin x = 6\]
Thus, the given equation is of the form
\[a\cos x + b\sin x = c\]
From the equation, we get
\[a = 3,b = 4\]and \[{\rm{c}} = 6\].
Let: \[a = 3 = r\cos \alpha \]and \[b = 4 = r\sin \alpha \]
Now, the expression using the values can be written as
\[\tan \alpha = \frac{b}{a} = \frac{4}{3}\]
Solve for\[\alpha \], by moving tan to the other side to calculate the value of\[\alpha \]:
\[ \Rightarrow \alpha = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)\]
Also, we know that
\[r = \sqrt {{a^2} + {b^2}} \]
Substitute the value of \[a\]and\[b\]:
\[ = \sqrt {9 + 16} \]
Add the terms inside the root to simplify the expression:
\[ = \sqrt {25} \]
Take square root of the obtained expression to solve the value for\[r\]:
\[r = 5\]
On putting the values of \[a = 3 = r\cos \alpha \] and \[b = 4 = r\sin \alpha \] in equation (i), we get:
\[r\cos \alpha \cos \theta + \sin \alpha \sin \theta = 6\]
This can be written as
\[ \Rightarrow r\cos (\theta - \alpha ) = 6\]
As both the roots are imaginary there exists no value of \[x\] satisfying this given equation.
Therefore, the equation \[3\cos x + 4\sin x = 6\] has no sol ution.
Hence, the option D is correct.
Note
Trigonometry and inequality ideas must be thoroughly understood to answer such queries. To solve the presented problem, one must be familiar with the methods for computing the range of trigonometric expressions of the form (\[asinx + bcosx\]). To solve such issues, we also need to understand how to calculate the range of a composite complex number.
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