Question

If \[2\cos \theta + \sin \theta = 1\],\[(\theta \ne \dfrac{\pi }{2})\], then \[7\cos \theta + 6\sin \theta \] is equal to
A. \[\dfrac{{11}}{2}\]
B. \[\dfrac{{46}}{5}\]
C. \[\dfrac{1}{2}\]
D. \[2\]

Answer 1

Hint: We use the first equation to write value one trigonometric function in term of other and then we square both sides to make use of the \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and assuming one function as a variable we write the other function in terms of variable. This will give us a quadratic equation, find the roots using factorization method.

Complete step-by-step answer:
We have \[2\cos \theta + \sin \theta = 1\]
Shifting the value of \[\sin \theta \] to right and side of the equation we get
\[ \Rightarrow 2\cos \theta = 1 - \sin \theta \]
Squaring both sides of the equation, we get
\[ \Rightarrow {\left( {2\cos \theta } \right)^2} = {\left( {1 - \sin \theta } \right)^2}\]
\[ \Rightarrow 4{\cos ^2}\theta = {\left( {1 - \sin \theta } \right)^2}\] … (1)
Let us assume the value of \[\sin \theta = x\]
We know \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Shifting \[{\sin ^2}\theta \]to right hand side of the equation
\[ \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta \]
Put \[{\sin ^2}\theta = {x^2}\]
\[ \Rightarrow {\cos ^2}\theta = 1 - {x^2}\] … (2)
Substitute the value of \[{\cos ^2}\theta = 1 - {x^2}\] from equation (2) and \[\sin \theta = x\] in equation (1)
\[ \Rightarrow 4(1 - {x^2}) = {\left( {1 - x} \right)^2}\]
Open the right hand side of the equation by using \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] where \[a = 1,b = x\].
\[ \Rightarrow 4 - 4{x^2} = 1 + {x^2} - 2x\]
Shift all the values to one side of the equation
\[
   \Rightarrow - 4{x^2} - {x^2} + 2x + 4 - 1 = 0 \\
   \Rightarrow - 5{x^2} + 2x + 3 = 0 \\
 \]
Taking negative sign common
\[ \Rightarrow 5{x^2} - 2x - 3 = 0\]
Now we can write \[ - 2x = - 5x + 3x\]
\[ \Rightarrow 5{x^2} - 5x + 3x - 3 = 0\]
Make factors
\[
   \Rightarrow 5x(x - 1) + 3(x - 1) = 0 \\
   \Rightarrow (5x + 3)(x - 1) = 0 \\
 \]
Now we equate both factors to zero
\[
  5x + 3 = 0 \\
  5x = - 3 \\
  x = \dfrac{{ - 3}}{5} \\
 \]
Now we substitute \[\sin \theta = x\]
\[\sin \theta = \dfrac{{ - 3}}{5}\]
Then from equation \[2\cos \theta + \sin \theta = 1\]
\[2\cos \theta + \dfrac{{ - 3}}{5} = 1\]
Shift all constants to one side
\[2\cos \theta = 1 + \dfrac{3}{5}\]
Take LCM on RHS of the equation
\[
  2\cos \theta = \dfrac{{5 + 3}}{5} \\
  2\cos \theta = \dfrac{8}{5} \\
 \]
Divide both sides by 2
\[\cos \theta = \dfrac{8}{5} \times \dfrac{1}{2}\]
\[\cos \theta = \dfrac{4}{5}\]
And from the second factor
\[
  x - 1 = 0 \\
  x = 1 \\
 \]
Now we substitute \[\sin \theta = x\]
\[
  \sin \theta = 1 \\
  \theta = \dfrac{\pi }{2} \\
 \]
Which is not acceptable as the condition in the question is \[(\theta \ne \dfrac{\pi }{2})\]
So, the value of \[\sin \theta = \dfrac{{ - 3}}{5}\] and \[\cos \theta = \dfrac{4}{5}\]
Substitute the values in \[7\cos \theta + 6\sin \theta \]
\[
   \Rightarrow 7\cos \theta + 6\sin \theta = 7 \times \dfrac{4}{5} + 6 \times \dfrac{{ - 3}}{5} \\
   \Rightarrow 7\cos \theta + 6\sin \theta = \dfrac{{28}}{5} + \dfrac{{ - 18}}{5} \\
 \]
Taking LCM on right hand side of the equation
\[
   \Rightarrow 7\cos \theta + 6\sin \theta = \dfrac{{28 - 18}}{5} \\
   \Rightarrow 7\cos \theta + 6\sin \theta = \dfrac{{10}}{5} \\
 \]
Cancel out same factors from both numerator and denominator
\[ \Rightarrow 7\cos \theta + 6\sin \theta = 2\]

So, option D is correct.

Note: Students make mistake of finding the value of \[\theta \] by taking inverse on both sides after we have obtained values of \[\sin \theta ,\cos \theta \] which is not required in the solution as RHS has direct substitution of values of \[\sin \theta ,\cos \theta \].
Students many times make mistake of directly writing one trigonometric function in term of other, like if \[2\cos \theta + \sin \theta = 1\] then, they will transform it into \[\sin \theta = 1 - 2\cos \theta \] and then they will substitute values in \[7\cos \theta + 6\sin \theta \] to get the answer which is wrong.