Question
Question: The equation \(\sqrt{3}\sin x+\cos x=4\) has A) Infinitely many solutions B) No solutions C) ...
The equation 3sinx+cosx=4 has
A) Infinitely many solutions
B) No solutions
C) Two solutions
D) Only one solution
Solution
For answering this question we should solve the given trigonometric expression 3sinx+cosx=4 . For doing that we will simplify and transform the given expression into the form of the basic trigonometric formulae sinAcosB+cosAsinB=sin(A+B).
Complete step by step solution:
Now considering from the question we have been asked to simplify the given trigonometric expression 3sinx+cosx=4 .
Now we will divide the whole expression by 2 on both sides. By doing that we will have 21(3sinx+cosx)=24⇒23sinx+21cosx=2
Now if we observe carefully then we can say that it is in the form of the basic trigonometric formulae sinAcosB+cosAsinB=sin(A+B) .
Now for further simplifying it we will use the values cos30∘=23 and sin30∘=21 from the basic trigonometric concepts.
Now we can write this expression as ⇒cos30∘sinx+sin30∘cosx=2 .
Now we will simplify and write it as ⇒sin(30∘+x)=2 .
From the basic concepts of trigonometry we know that the range of sine trigonometric function is given as −1≤sinθ≤1 .
Since the value in the right hand side of the trigonometric expression is greater than one. So there does not exist any values for x .
Hence we can conclude that the number of solutions of the given trigonometric expression 3sinx+cosx=4 is zero. So we need to mark the option B as correct.
So, the correct answer is “Option B”.
Note: While answering questions of this type we should be sure with our trigonometric concepts. This question can be answered easily and in a short span of time and very few mistakes are possible in it. We have many other trigonometric formulae similarly for example some of them are sinAcosB−cosAsinB=sin(A−B) , cosAcosB−sinAsinB=cos(A+B) and cosAcosB+sinAsinB=cos(A−B) .