Question

If $2 \sin\theta+3 \cos\theta=2$ and $\theta \neq (2n+1)\frac{\pi}{2}$ then $\sin\theta+\cos\theta=$

• A. $5/13$
• B. $3/5$
• C. $7/13$
• D. $4/5$

Hint:

Equations of the form $a\sin\theta + b\cos\theta = c$ can be solved effectively using the tangent half-angle substitutions: $\sin\theta = \frac{2t}{1+t^2}$ and $\cos\theta = \frac{1-t^2}{1+t^2}$ where $t = \tan(\theta/2)$. This transforms the trigonometric equation into a quadratic equation in $t$.

Answer 1

The Correct Option is C

We are given the equation $2 \sin\theta+3 \cos\theta=2$.
We use the tangent half-angle substitution, where $t = \tan(\theta/2)$.
$\sin\theta = \frac{2t}{1+t^2}$ and $\cos\theta = \frac{1-t^2}{1+t^2}$.
Substituting these into the given equation:
$2\left(\frac{2t}{1+t^2}\right) + 3\left(\frac{1-t^2}{1+t^2}\right) = 2$.
Multiply by $(1+t^2)$:
$4t + 3(1-t^2) = 2(1+t^2)$.
$4t + 3 - 3t^2 = 2 + 2t^2$.
Rearranging the terms to form a quadratic equation:
$5t^2 - 4t - 1 = 0$.
Factoring the quadratic equation:
$5t^2 - 5t + t - 1 = 0 \implies 5t(t-1) + 1(t-1) = 0 \implies (5t+1)(t-1) = 0$.
This gives two possible values for $t$: $t=1$ or $t=-1/5$.
If $t = \tan(\theta/2) = 1$, then $\theta/2 = n\pi + \pi/4$, which gives $\theta = 2n\pi + \pi/2$. This corresponds to a case where $\cos\theta=0$. The condition $\theta \neq (2n+1)\frac{\pi}{2}$ might be intended to exclude this, although $\theta = \pi/2$ is a solution to the original equation. Let's evaluate for the other case.
If $t = \tan(\theta/2) = -1/5$:
$\sin\theta = \frac{2(-1/5)}{1+(-1/5)^2} = \frac{-2/5}{1+1/25} = \frac{-2/5}{26/25} = -\frac{2}{5} \times \frac{25}{26} = -\frac{5}{13}$.
$\cos\theta = \frac{1-(-1/5)^2}{1+(-1/5)^2} = \frac{1-1/25}{1+1/25} = \frac{24/25}{26/25} = \frac{24}{26} = \frac{12}{13}$.
Now, we find the value of $\sin\theta + \cos\theta$.
$\sin\theta + \cos\theta = -\frac{5}{13} + \frac{12}{13} = \frac{7}{13}$.