Question
Question: How do you solve \(\sin x - 2\cos x = 2?\)...
How do you solve sinx−2cosx=2?
Solution
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)=sinacosb−sinbcosa
Use the above information to solve the problem.
Formula used:
Compound angle formula of sine, sin(a−b)=sinacosb−sinbcosa
Complete step by step solution:
In order to solve the given trigonometric equation sinx−2cosx=2
we will first express the coefficient of cosine in terms of tangent,
That is let us consider,
tany=2 ∴y=tan−1(2)=63.430
We can write the given trigonometric equation as
⇒sinx−2cosx=2 ⇒sinx−tanycosx=2
Now expressing tangent as the ratio of sine is to cosine, we will get
⇒sinx−cosysinycosx=2 ⇒sinxcosy−sinycosx=2cosy
Using compound angle formula of sine function to solve it further, we will get
⇒sinxcosy−sinycosx=2cosy ⇒sin(x−y)=2cosy
Now putting the value of y and also substituting the value of cosy we will get,
⇒sin(x−63.430)=2cos(63.430) ⇒sin(x−63.430)=2×0.45[∵cos(63.430)=0.45] ⇒sin(x−63.430)=0.90
Taking inverse function of sine both sides we will get,
⇒sin−1(sin(x−63.430))=sin−1(0.90) ⇒x−63.430=sin−1(0.90)
Now, sin−1(0.90) will give two values in the principle interval i.e. (0,3600)
One of the two solutions is given as
⇒x−63.430=sin−1(0.90) ⇒x−63.430=64.160 ⇒x=63.430+64.160≈1280
Another one is given as
Therefore we get two solutions for the trigonometric equation sinx−2cosx=2 in the principle interval.
Note: We have taken both the solutions, because the sine function shows positive sign twice in between the interval (0,3600) 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.