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Question: How do you solve \(\sin 2\theta = \cos \theta \) ?...

How do you solve sin⁡2θ=cos⁡θ\sin 2\theta = \cos \theta ?

Explanation

Solution

1.1. Start this question by using the Double angle trigonometric formula for sine.
2.2. Expand the formula and simplify by adding or subtracting other trigonometric terms.
3.3. Simplify until it cannot simplify further and until it equals to 0.

Formula used:
we are going to use Double angle trigonometric formula for sine:
⇒sin⁡2θ=2sin⁡θ×cos⁡θ\Rightarrow \sin 2\theta = 2\sin \theta \times \cos \theta

Complete step by step answer:
Firstly, we will be using the double angle formula for sine on the expression given to us:
⇒sin⁡2θ=cos⁡θ\Rightarrow \sin 2\theta = \cos \theta
After applying the formula we get:
⇒2sin⁡θ×cos⁡θ=cos⁡θ\Rightarrow 2\sin \theta \times \cos \theta = \cos \theta
Subtract with cos⁡θ\cos \theta on both the sides of the equation above we get:
⇒2sin⁡θ×cos⁡θ−cos⁡θ=cos⁡θ−cos⁡θ\Rightarrow 2\sin \theta \times \cos \theta - \cos \theta = \cos \theta - \cos \theta
Simplify and rewrite the equation:
⇒2sin⁡θ×cos⁡θ−cos⁡θ=0\Rightarrow 2\sin \theta \times \cos \theta - \cos \theta = 0
Take out the common factor of cos⁡θ\cos \theta from left hand side of the equation we get:
⇒cos⁡θ(sin⁡θ−1)=0\Rightarrow \cos \theta \left( {\sin \theta - 1} \right) = 0
Now, we will evaluate both the terms of left hand side equals to zero and solve them separately:
Equating cos⁡θ\cos \theta equals to zero we get:
⇒cos⁡θ=0\Rightarrow \cos \theta = 0
From the above expression we can now find the value of θ\theta the value of cos become zero when:
⇒θ=π2,3π2\Rightarrow \theta = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}……………………. Eq.(22)
Similarly Equating 2sin⁡θ−12\sin \theta - 1equals to zero we get:
⇒2sin⁡θ−1=0\Rightarrow 2\sin \theta - 1 = 0
Adding 11 to both sides of the equation:
⇒2sin⁡θ−1+1=0+1\Rightarrow 2\sin \theta - 1 + 1 = 0 + 1
Simplify and rewrite:
⇒2sin⁡θ=1\Rightarrow 2\sin \theta = 1
Divided by 22 on both the sides of the equation:
⇒2sin⁡θ2=12\Rightarrow \dfrac{{2\sin \theta }}{2} = \dfrac{1}{2}
Simplify and rewrite:
⇒2sin⁡θ2=12\Rightarrow \dfrac{{{2}\sin \theta }}{{{2}}} = \dfrac{1}{2}
After cancelling we get,
⇒sin⁡θ=12\Rightarrow \sin \theta = \dfrac{1}{2}
From the above expression we can now find the value of θ\theta the value of sin become 1/2 when:
⇒θ=π6,5π6\Rightarrow \theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}………………………… eq. (2)(2)
From equation 1 and 2 we get four solution π2,3π2,π6,5π6\dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{\pi }{6},\dfrac{{5\pi }}{6}within the range 00 to 2π2\pi .

Note: Remember that while solving these, you should only change one side of the equation and expand it further.
Before proceeding to a solution, it's important to know the double-angle identity for cosines.
There are three formulas, but since both sides contain sine, we're going to use the formula that includes only sines.
The formula is sin⁡2θ=2sin⁡θ×cos⁡θ\sin 2\theta = 2\sin \theta \times \cos \theta .