Solveeit Logo
Login

Question

Question: Solve \[3\sin x + 4\cos x = 5 \Rightarrow 6\tan \dfrac{x}{2} - 9{\tan ^2}\dfrac{x}{2}\]...

Solve 3sin⁡x+4cos⁡x=5⇒6tan⁡x2−9tan⁡2x23\sin x + 4\cos x = 5 \Rightarrow 6\tan \dfrac{x}{2} - 9{\tan ^2}\dfrac{x}{2}

Explanation

Solution

In this question, we have to find the value of 6tan⁡x2−9tan⁡2x26\tan \dfrac{x}{2} - 9{\tan ^2}\dfrac{x}{2}.
We will solve the above question with the help of the given expression 3sin⁡x+4cos⁡x=53\sin x + 4\cos x = 5in the question.
To find the value, first, we will find the value oftan⁡x2\tan \dfrac{x}{2} from the given expression3sin⁡x+4cos⁡x=53\sin x + 4\cos x = 5and we will put the value of tan⁡x2\tan \dfrac{x}{2}in 6tan⁡x2−9tan⁡2x26\tan \dfrac{x}{2} - 9{\tan ^2}\dfrac{x}{2}.
FORMULA TO BE USED: sin⁡x=2tan⁡x21+tan⁡2x2\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}} and cos⁡x=1−tan⁡2x21+tan⁡2x2\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}

Complete answer:
In this question, the expression is given as: 3\sin x + 4\cos x = 5$$$$...........(i)
Using the formula sin⁡x=2tan⁡x21+tan⁡2x2\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}and cos⁡x=1−tan⁡2x21+tan⁡2x2\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}
Now putting the formula in eq.(i)(i), we get,
32tan⁡x21+tan⁡2x2+41−tan⁡2x21+tan⁡2x2=53\dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}} + 4\dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}} = 5
Now, multiply and add the value, we get,
6tan⁡x2+4−4tan⁡2x21+tan⁡2x2=5\dfrac{{6\tan \dfrac{x}{2} + 4 - 4{{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}} = 5
Now transfer the denominator part to RHS and solve, we get,
6tan⁡x2+4−4tan⁡2x2=5+5tan⁡2x26\tan \dfrac{x}{2} + 4 - 4{\tan ^2}\dfrac{x}{2} = 5 + 5{\tan ^2}\dfrac{x}{2}
Now transfer all the value of RHS to LHS, we get,
6tan⁡x2+4−4tan⁡2x2−5−5tan⁡2x2=06\tan \dfrac{x}{2} + 4 - 4{\tan ^2}\dfrac{x}{2} - 5 - 5{\tan ^2}\dfrac{x}{2} = 0
Now, add all the like terms, we get,
\Rightarrow $$$$9{\tan ^2}\dfrac{x}{2} - 6\tan \dfrac{x}{2} + 1 = 0
\Rightarrow $$$${(3\tan \dfrac{x}{2})^2} - 2 \times 3\tan \dfrac{x}{2} + 1 = 0$$$$...........(ii)
Here we see that it is looking like the identity (a2−2ab+b2)=(a−b)2({a^2} - 2ab + {b^2}) = {(a - b)^2}
So, we will apply this identity in eq. (ii)(ii)we get,
\Rightarrow $$$${(3\tan \dfrac{x}{2} - 1)^2} = 0
\Rightarrow $$$$3\tan \dfrac{x}{2} = 1
\Rightarrow $$$$\tan \dfrac{x}{2} = \dfrac{1}{3}$$$$...........(iii)
Now, put the value of eq. (iii)(iii) in 6tan⁡x2−9tan⁡2x26\tan \dfrac{x}{2} - 9{\tan ^2}\dfrac{x}{2},we get,
=6×13−9(13)2= 6 \times \dfrac{1}{3} - 9{(\dfrac{1}{3})^2}
=6×13−9(19)= 6 \times \dfrac{1}{3} - 9(\dfrac{1}{9})
=2−1= 2 - 1
So, the answer to the expression 6tan⁡x2−9tan⁡2x26\tan \dfrac{x}{2} - 9{\tan ^2}\dfrac{x}{2} is 11.

Note:
Always in such type of question i.e., whenever some trigonometric expression is given as different type, along with their equation value and you have to find the solution of other expression using the given one, always try to solve the first expression to get values of the terms and put them in expression which you have to find.
Before solving any trigonometric expression first learn all the formulas to apply whenever required in that type of question.
Always try to solve the first expression in the term which the second expression contains which you have to find.