Question

How do you solve $\sin x - 2\cos x = 2?$

Answer 1

Hint: First express $2$ in terms of tangent of an argument and find the value of respective argument then convert tangent into sine and cosine and simplify further using the compound angle formula of sine and finally find the required solution in the principal interval. Compound angle formula of sine function is given as:
$\sin (a - b) = \sin a\cos b - \sin b\cos a$
Use the above information to solve the problem.

Formula used:
 Compound angle formula of sine, $\sin (a - b) = \sin a\cos b - \sin b\cos a$

Complete step by step solution:
In order to solve the given trigonometric equation $\sin x - 2\cos x = 2$
we will first express the coefficient of cosine in terms of tangent,
That is let us consider,
$
\tan y = 2 \\
\therefore y = {\tan ^{ - 1}}(2) = {63.43^0} \\
$
We can write the given trigonometric equation as
$
\Rightarrow \sin x - 2\cos x = 2 \\
\Rightarrow \sin x - \tan y\cos x = 2 \\
$
Now expressing tangent as the ratio of sine is to cosine, we will get
$
\Rightarrow \sin x - \dfrac{{\sin y}}{{\cos y}}\cos x = 2 \\
\Rightarrow \sin x\cos y - \sin y\cos x = 2\cos y \\
$
Using compound angle formula of sine function to solve it further, we will get
$
\Rightarrow \sin x\cos y - \sin y\cos x = 2\cos y \\
\Rightarrow \sin (x - y) = 2\cos y \\
$
Now putting the value of $y$ and also substituting the value of $\cos y$ we will get,
$
\Rightarrow \sin \left( {x - {{63.43}^0}} \right) = 2\cos \left( {{{63.43}^0}} \right) \\
\Rightarrow \sin \left( {x - {{63.43}^0}} \right) = 2 \times 0.45\;\;\;\;\;\left[ {\because \cos \left(
{{{63.43}^0}} \right) = 0.45} \right] \\
\Rightarrow \sin \left( {x - {{63.43}^0}} \right) = 0.90 \\
$
Taking inverse function of sine both sides we will get,
$
\Rightarrow {\sin ^{ - 1}}\left( {\sin \left( {x - {{63.43}^0}} \right)} \right) = {\sin ^{ - 1}}\left( {0.90}
\right) \\
\Rightarrow x - {63.43^0} = {\sin ^{ - 1}}\left( {0.90} \right) \\
$
Now, ${\sin ^{ - 1}}\left( {0.90} \right)$ will give two values in the principle interval i.e. $\left(
{0,\;{{360}^0}} \right)$
One of the two solutions is given as
$
\Rightarrow x - {63.43^0} = {\sin ^{ - 1}}\left( {0.90} \right) \\
\Rightarrow x - {63.43^0} = {64.16^0} \\
\Rightarrow x = {63.43^0} + {64.16^0} \approx {128^0} \\
$
Another one is given as
\[
\Rightarrow x - {63.43^0} = {\sin ^{ - 1}}\left( {0.90} \right) \\

\Rightarrow x - {63.43^0} = {180^0} - {64.16^0} \\
\Rightarrow x = {115.84^0} + {63.43^0} = {179.27^0} \\
\]
Therefore we get two solutions for the trigonometric equation $\sin x - 2\cos x = 2$ in the principle interval.

Note: We have taken both the solutions, because the sine function shows positive sign twice in between the interval $\left( {0,\;{{360}^0}} \right)$ that is in the first quadrant and in the second quadrant. You can check both the solutions by substituting it in the function. Also, when finding the inverse value of a trigonometric function then always considers its principle solution.