
Find the value of x if \[2\sin 3x=\sqrt{3}\].
Answer
607.8k+ views
Hint: We will first divide the given expression on both sides to transform it into a recognizable form. Then we will check for which standard angle it has this value and then we will substitute and will do the arithmetic to arrive at the answer.
Complete step-by-step answer:
The expression mentioned in the question is \[2\sin 3x=\sqrt{3}..........(1)\]
Now dividing both sides of equation (1) by 2 we get,
\[\Rightarrow \dfrac{2\sin 3x}{2}=\dfrac{\sqrt{3}}{2}..........(2)\]
Now cancelling the similar terms in equation (2) we get,
\[\Rightarrow \sin 3x=\dfrac{\sqrt{3}}{2}..........(3)\]
Also we know that sin 60 is equal to \[\dfrac{\sqrt{3}}{2}\] and hence substituting it in place of \[\dfrac{\sqrt{3}}{2}\] in equation (3) we get,
\[\Rightarrow \sin 3x=\sin {{60}^{\circ }}..........(4)\]
Now sin gets cancelled from both sides in equation (4) and hence we get,
\[\Rightarrow 3x={{60}^{\circ }}..........(5)\]
Now isolating x in equation (5) by dividing both sides by 3 and hence we get,
\[\Rightarrow \dfrac{3x}{3}=\dfrac{{{60}^{\circ }}}{3}..........(6)\]
Hence cancelling similar terms in the left hand side of equation (6) and hence we get on solving,
\[\Rightarrow x={{20}^{\circ }}\]
Hence the value of x is \[{{20}^{\circ }}\].
Note: We have to be very careful with the value of standard angles of sin, cos and tan because it is the key here. Also we in a hurry can make a mistake in solving equation (5) as we can miss writing 3. Also we should not forget to write the degree next to the number because then it will be wrong.
Complete step-by-step answer:
The expression mentioned in the question is \[2\sin 3x=\sqrt{3}..........(1)\]
Now dividing both sides of equation (1) by 2 we get,
\[\Rightarrow \dfrac{2\sin 3x}{2}=\dfrac{\sqrt{3}}{2}..........(2)\]
Now cancelling the similar terms in equation (2) we get,
\[\Rightarrow \sin 3x=\dfrac{\sqrt{3}}{2}..........(3)\]
Also we know that sin 60 is equal to \[\dfrac{\sqrt{3}}{2}\] and hence substituting it in place of \[\dfrac{\sqrt{3}}{2}\] in equation (3) we get,
\[\Rightarrow \sin 3x=\sin {{60}^{\circ }}..........(4)\]
Now sin gets cancelled from both sides in equation (4) and hence we get,
\[\Rightarrow 3x={{60}^{\circ }}..........(5)\]
Now isolating x in equation (5) by dividing both sides by 3 and hence we get,
\[\Rightarrow \dfrac{3x}{3}=\dfrac{{{60}^{\circ }}}{3}..........(6)\]
Hence cancelling similar terms in the left hand side of equation (6) and hence we get on solving,
\[\Rightarrow x={{20}^{\circ }}\]
Hence the value of x is \[{{20}^{\circ }}\].
Note: We have to be very careful with the value of standard angles of sin, cos and tan because it is the key here. Also we in a hurry can make a mistake in solving equation (5) as we can miss writing 3. Also we should not forget to write the degree next to the number because then it will be wrong.
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