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Question: Find the possible values of \(\sin x\), if \(8\sin x - \cos x = 4\)....

Find the possible values of sin⁡x\sin x, if 8sin⁡x−cos⁡x=48\sin x - \cos x = 4.

Explanation

Solution

To solve this question, we will use some basic trigonometric identities. By using the identity, sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1, we can simplify the given equation and then we can get the value of sin x.

Complete step-by-step answer:
Given,
8sin⁡x−cos⁡x=48\sin x - \cos x = 4 ………………….……….. (i)
We have to find out all the possible values of sin x.
So,
Form equation (i),
⇒8sin⁡x−cos⁡x=4\Rightarrow 8\sin x - \cos x = 4
⇒8sin⁡x−4=cos⁡x\Rightarrow 8\sin x - 4 = \cos x ……………. ………. (ii)
We know that,
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1.
Putting the value of cos x from equation (ii), we will get
⇒sin⁡2x+(8sin⁡x−4)2=1\Rightarrow {\sin ^2}x + {\left( {8\sin x - 4} \right)^2} = 1.
Solving this by using the identity, (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, we will get
⇒sin⁡2x+64sin⁡2x+16−64sin⁡x=1 ⇒65sin⁡2x−64sin⁡x+15=0  \Rightarrow {\sin ^2}x + 64{\sin ^2}x + 16 - 64\sin x = 1 \\\ \Rightarrow 65{\sin ^2}x - 64\sin x + 15 = 0 \\\
Solving the above quadratic equation by splitting the middle term, we will get

⇒65sin⁡2x−(39+25)sin⁡x+15=0 ⇒65sin⁡2x−39sin⁡x−25sin⁡x+15=0 ⇒13sin⁡x(5sin⁡x−3)−5(5sin⁡x−3)=0 ⇒(13sin⁡x−5)(5sin⁡x−3)=0  \Rightarrow 65{\sin ^2}x - \left( {39 + 25} \right)\sin x + 15 = 0 \\\ \Rightarrow 65{\sin ^2}x - 39\sin x - 25\sin x + 15 = 0 \\\ \Rightarrow 13\sin x\left( {5\sin x - 3} \right) - 5\left( {5\sin x - 3} \right) = 0 \\\ \Rightarrow \left( {13\sin x - 5} \right)\left( {5\sin x - 3} \right) = 0 \\\

Therefore,
(5sin⁡x−3)=0 ⇒5sin⁡x=3  \left( {5\sin x - 3} \right) = 0 \\\ \Rightarrow 5\sin x = 3 \\\
sin⁡x=35\sin x = \dfrac{3}{5}
And,

⇒(13sin⁡x−5)=0 ⇒13sin⁡x=5 sin⁡x=513 \Rightarrow \left( {13\sin x - 5} \right) = 0 \\\ \Rightarrow 13\sin x = 5 \\\ \sin x = \dfrac{5}{{13}} \\\

Hence, the possible values of sin⁡x\sin x in 8sin⁡x−cos⁡x=48\sin x - \cos x = 4 are sin⁡x=(35,513)\sin x = \left( {\dfrac{3}{5},\dfrac{5}{{13}}} \right).

Note: Whenever we ask such types of questions, we have to use some basic trigonometric identities. First, we have to write the given equation in a simplified form and then according to that form, we will use the suitable identity which is useful. After that by using the identity, we are now able to make a quadratic equation in terms of trigonometric function. Now, we can solve that quadratic equation easily and by solving it, we will get the required possible answers.