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

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.