Question

How do you solve \[\cos t-\sin (2t)=0\]?

Answer 1

Hint: As we can see that this question is from the topic of trigonometry. Therefore, we should have a better knowledge in trigonometry. We should know the formulas of trigonometry chapters like \[\sin 2x=2\sin x\cos x\] as we are going to use here.

Complete step-by-step solution:
Let us solve this question.
So, in this question we have asked to solve
\[\cos t-\sin \left( 2t \right)=0\]
Using the formula \[\sin 2x=2\sin x\cos x\] in the above equation, we get
\[\Rightarrow \cos t-2\sin t\cos t=0\]
Taking common \[\cos t\] to the both sides of the equation, we get
\[\Rightarrow \left( \cos t \right)\left( 1-2\sin t \right)=0\]
In the above equation, two terms are multiplied and both the terms are in the parenthesis which is equal to zero. So, we can say from the above equation that
\[\Rightarrow \cos t=0\] and \[1-2\sin t=0\].
Hence, we get from the equation \[1-2\sin t=0\] that
\[2\sin t=1\]
We can say that the above equation can also be written as
\[\sin t=\dfrac{1}{2}\]
So, now we can say that we have to solve the two equations that are \[\cos t=0\] and \[\sin t=\dfrac{1}{2}\].
As we know that,
When \[\cos t=0\], then the value of t will be \[\dfrac{\pi }{2}\] and \[\dfrac{3\pi }{2}\] in the range of \[\left[ 0,2\pi \right]\]
And when \[\sin t=\dfrac{1}{2}\], then the value of t will be \[\dfrac{\pi }{6}\] and \[\dfrac{5\pi }{6}\] in the range of \[\left[ 0,2\pi \right]\].
So, we have solved the equation now. And, we have got the values as \[\dfrac{\pi }{6}\], \[\dfrac{\pi }{2}\], \[\dfrac{5\pi }{6}\], and \[\dfrac{3\pi }{2}\].

Note: For solving these types of questions, formulas, identities, values should be kept remembered and should be capable of finding the angles if the values are given. We should remember the formulas below so that it will be easy to find the solution for these types of questions.
When \[\sin t=0\] then the value of t will be \[0\], \[\pi \], and \[2\pi \]
When \[\sin t=\dfrac{1}{2}\] then the value of t will be \[\dfrac{\pi }{6}\], and\[\dfrac{5\pi }{6}\]
When \[\sin t=\dfrac{1}{\sqrt{2}}\] then the value of t will be \[\dfrac{\pi }{4}\], and \[\dfrac{5\pi }{4}\]
When \[\sin t=\dfrac{\sqrt{3}}{2}\] then the value of t will be \[\dfrac{\pi }{3}\], and \[\dfrac{2\pi }{3}\]
When \[\sin t=1\] then the value of t will be \[\dfrac{\pi }{2}\]
When \[\cos t=1\] then the value of t will be \[0\], and \[2\pi \]
When \[\cos t=\dfrac{\sqrt{3}}{2}\] then the value of t will be \[\dfrac{\pi }{6}\], and \[\dfrac{11\pi }{6}\]
When \[\cos t=\dfrac{1}{\sqrt{2}}\] then the value of t will be \[\dfrac{\pi }{4}\], and \[\dfrac{7\pi }{4}\]
When \[\cos t=\dfrac{1}{2}\] then the value of t will be \[\dfrac{\pi }{3}\], and \[\dfrac{5\pi }{3}\]
When \[\cos t=0\] then the value of t will be \[\dfrac{\pi }{2}\], and \[\dfrac{3\pi }{2}\]
All the value of t in the above formula is for the range of \[\left[ 0,2\pi \right]\].