Answer
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Hint: Here we will use the formula of the sum of cosine angles to simplify the equation. Then we will apply a sine function for double angle and simplify the equation further. We will then equate one of the products that will be obtained after simplifying to the value on the right-hand side of the equation. Using this we will find the solution to the given equation.
Complete step-by-step answer:
The equation given is \[\cos x + 2\sin 2x + \cos 3x = 3\].
No using the formula \[\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\] in the above equation, we get
\[ \Rightarrow 2\sin 2x + 2\cos \left( {\dfrac{{3x + x}}{2}} \right)\cos \left( {\dfrac{{3x - x}}{2}} \right) = 3\]
Adding and subtracting the terms in the denominator, we get
\[ \Rightarrow 2\sin 2x + 2\cos 2x\cos x = 3\]
Now simplifying sin2x term by using formula \[\sin 2x = 2\sin x\cos x\], we get
\[ \Rightarrow 2 \times 2\sin x\cos x + 2\cos 2x\cos x = 3\]
Taking 2cosx common from both the terms, we get
\[ \Rightarrow 2\cos x\left( {\cos 2x + 2\sin x} \right) = 3\]
Now we can state that,
\[\begin{array}{l}2\cos x = 3\\ \Rightarrow \cos x = \dfrac{3}{2}\end{array}\]
We know for \[x \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)\], so \[cosx\] can’t have value as \[\dfrac{3}{2}\].
There is no need to check the term in the bracket as above one is not satisfying the condition
So, this means the equation has no solution.
Hence, the option (D) is correct.
Note: This question is solved by using the hit and trial method, where we simplified the term and check whether we are getting the term in the product or not. Trigonometric functions that are used in modern mathematics are the sine, cosine, and tangent. There are many identities of the trigonometric functions such as Pythagorean identity, sum, and difference formulas and derivatives and anti-derivatives. Trigonometric identities are only used in equations where trigonometric functions are present.
Complete step-by-step answer:
The equation given is \[\cos x + 2\sin 2x + \cos 3x = 3\].
No using the formula \[\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\] in the above equation, we get
\[ \Rightarrow 2\sin 2x + 2\cos \left( {\dfrac{{3x + x}}{2}} \right)\cos \left( {\dfrac{{3x - x}}{2}} \right) = 3\]
Adding and subtracting the terms in the denominator, we get
\[ \Rightarrow 2\sin 2x + 2\cos 2x\cos x = 3\]
Now simplifying sin2x term by using formula \[\sin 2x = 2\sin x\cos x\], we get
\[ \Rightarrow 2 \times 2\sin x\cos x + 2\cos 2x\cos x = 3\]
Taking 2cosx common from both the terms, we get
\[ \Rightarrow 2\cos x\left( {\cos 2x + 2\sin x} \right) = 3\]
Now we can state that,
\[\begin{array}{l}2\cos x = 3\\ \Rightarrow \cos x = \dfrac{3}{2}\end{array}\]
We know for \[x \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)\], so \[cosx\] can’t have value as \[\dfrac{3}{2}\].
There is no need to check the term in the bracket as above one is not satisfying the condition
So, this means the equation has no solution.
Hence, the option (D) is correct.
Note: This question is solved by using the hit and trial method, where we simplified the term and check whether we are getting the term in the product or not. Trigonometric functions that are used in modern mathematics are the sine, cosine, and tangent. There are many identities of the trigonometric functions such as Pythagorean identity, sum, and difference formulas and derivatives and anti-derivatives. Trigonometric identities are only used in equations where trigonometric functions are present.
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