Question

If $2 \cosx + \sin x = 1$, then find the sum of all possible values of $7 \cosx + 6 \sin x$.

Answer 1

8

Answer 2

Let $\cos x = c$ and $\sin x = s$.

We are given the equation: $2c + s = 1$

We need to find the sum of all possible values of $E = 7c + 6s$.

From the given equation, we can express $s$ in terms of $c$: $s = 1 - 2c$

Substitute this expression for $s$ into the fundamental trigonometric identity $c^2 + s^2 = 1$: $c^2 + (1 - 2c)^2 = 1$ $c^2 + (1 - 4c + 4c^2) = 1$ $5c^2 - 4c + 1 = 1$ $5c^2 - 4c = 0$ Factor out $c$: $c(5c - 4) = 0$

This equation gives two possible values for $c$:

Case 1: $c = 0$ If $\cos x = 0$, then substitute this back into $s = 1 - 2c$: $s = 1 - 2(0) = 1$ So, $\sin x = 1$. This pair $(\cos x, \sin x) = (0, 1)$ satisfies $c^2 + s^2 = 0^2 + 1^2 = 1$. Now, calculate $E$ for this case: $E_1 = 7c + 6s = 7(0) + 6(1) = 6$.

Case 2: $5c - 4 = 0 \implies c = \frac{4}{5}$ If $\cos x = \frac{4}{5}$, then substitute this back into $s = 1 - 2c$: $s = 1 - 2\left(\frac{4}{5}\right) = 1 - \frac{8}{5} = -\frac{3}{5}$ So, $\sin x = -\frac{3}{5}$. This pair $(\cos x, \sin x) = \left(\frac{4}{5}, -\frac{3}{5}\right)$ satisfies $c^2 + s^2 = \left(\frac{4}{5}\right)^2 + \left(-\frac{3}{5}\right)^2 = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1$. Now, calculate $E$ for this case: $E_2 = 7c + 6s = 7\left(\frac{4}{5}\right) + 6\left(-\frac{3}{5}\right) = \frac{28}{5} - \frac{18}{5} = \frac{10}{5} = 2$.

The possible values for $7 \cos x + 6 \sin x$ are 6 and 2. The sum of all possible values is $6 + 2 = 8$.

Therefore, the sum of all possible values is 8.