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Question

Question: How do you solve \[2\sin t\cos t - \cos t = 0\] ?...

How do you solve 2sintcostcost=02\sin t\cos t - \cos t = 0 ?

Explanation

Solution

Hint : To solve this question, we first need to simplify the first term into a single trigonometric function, then transpose the other function to the other side of the equation. Next, we need to use another identity to convert one side into the same trigonometric function as the other, and finally, compare both the angles to find out the value of the variable.

Complete step by step solution:
To solve this equation, we will use the following identity: sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta .
On applying this identity, we can simplify the expression significantly, the result being:
sin2tcost=0\sin 2t - \cos t = 0
Now, shifting cost\cos t to the right-hand side of the equation, we obtain the following:
sin2t=cost\sin 2t = \cos t
Further, we will use the identity sinθ=cos(90θ)\sin \theta = \cos ({90^ \circ } - \theta ) to convert both terms into the same trigonometric function.
We have,
sin2t=cos(902t)=cost\sin 2t = \cos ({90^ \circ } - 2t) = \cos t
Now, by comparing, we obtain 902t=t{90^ \circ } - 2t = t .
Shifting 2t2t to the other side,

90=3t t=30  \Rightarrow {90^ \circ } = 3t \\\ \Rightarrow t = {30^ \circ } \;

Thus, we obtain the final answer to 2sintcostcost=02\sin t\cos t - \cos t = 0 as t=30t = {30^ \circ } .
So, the correct answer is “ t=30t = {30^ \circ } ”.

Note : Trigonometric identities can be applied only on right angles, because the trigonometric functions used in these identities were generated for right angled triangles only. These identities help us to reach from the given expression to the desired result quicker, and also help us in solving such equations.
Similar to the identity for the sine of the double of an angle, we have an identity for the cosine of the double of an angle. The identity goes as follows:
cos2θ=cos2θsin2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta ,
where θ\theta is the angle for which the identity is being applied.
These identities are also used to derive further identities.