Solveeit Logo
Login

Question

Question: How do you solve \(\sin 2\theta + \cos \theta = 0\)?...

How do you solve sin⁡2θ+cos⁡θ=0\sin 2\theta + \cos \theta = 0?

Explanation

Solution

Expand sin⁡2θ\sin 2\theta to simplify the equation and then take out common terms to solve the equationSince the question comprises of two unlike terms we will first try to either convert them into like terms or at least convert it to a more simpler equation. To do this, we will first simplify sin⁡2θ\sin 2\theta with 2sin⁡θcos⁡2\sin \theta \cos . After that, we will take cos⁡θ\cos \theta common from both the terms which will leave us either with cos⁡θ=0\cos \theta = 0or 2sin⁡θ+1=02\sin \theta + 1 = 0. Solving these two will give us two different solutions.

Complete step by step solution:
Here, the given equation is sin⁡2θ+cos⁡θ=0\sin 2\theta + \cos \theta = 0 (1)
But we know that,
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta (2)
Putting value of sin⁡2θ\sin 2\theta from equation 2 in 1 we get,
2sin⁡θcos⁡θ+cos⁡θ=02\sin \theta \cos \theta + \cos \theta = 0
Taking cos⁡θ\cos \theta common from the equation
cos⁡θ(2sin⁡θ+1)=0\cos \theta (2\sin \theta + 1) = 0
Now, we know that whenever the product of any two variables is zero. Either or both of the variable will be equal to zero
∴  \therefore \,\,either cos⁡θ=0\cos \theta = 0or 2sin⁡θ+1=02\sin \theta + 1 = 0
solving these two separately :-

cos⁡θ=0 ⇒θ=90∘ \cos \theta = 0 \\\ \Rightarrow \theta = {90^ \circ } \\\

and,
2sin⁡θ+1=0 ⇒2sin⁡θ=−1 ⇒sin⁡θ=−12 ⇒θ=7π6  2\sin \theta + 1 = 0 \\\ \Rightarrow 2\sin \theta = - 1 \\\ \Rightarrow \sin \theta = \dfrac{{ - 1}}{2} \\\ \Rightarrow \theta = \dfrac{{7\pi }}{6} \\\
Hence, for the given equation sin⁡2θ+cos⁡θ=0\sin 2\theta + \cos \theta = 0, we have 2 solution, θ=90∘\theta = {90^ \circ }or θ=7π6\theta = \dfrac{{7\pi }}{6}
Alternate solution:
You can also find the general solution of the two equation to make the answer more general
Here,
cos⁡θ=0\cos \theta = 0
Now general solution for cos⁡θ=0\cos \theta = 0is given by the formula θ=(2n+1)π2\theta = \dfrac{{(2n + 1)\pi }}{2}
and for ,
2sin⁡θ+1=0 ⇒2sin⁡θ=−1 ⇒sin⁡θ=−12 ⇒θ=7π6  2\sin \theta + 1 = 0 \\\ \Rightarrow 2\sin \theta = - 1 \\\ \Rightarrow \sin \theta = \dfrac{{ - 1}}{2} \\\ \Rightarrow \theta = \dfrac{{7\pi }}{6} \\\
Now general solution of sin⁡θ=−k\sin \theta = - k is given by θ=nπ+(−1)nk\theta = n\pi + {( - 1)^n}k
therefore, our general solution becomes sin⁡θ=−12\sin \theta = \dfrac{{ - 1}}{2} θ=nπ+(−1)n(7π6)\theta = n\pi + {( - 1)^n}(\dfrac{{7\pi }}{6})
Hence, two general solution of sin⁡2θ+cos⁡θ=0\sin 2\theta + \cos \theta = 0are θ=(2n+1)π2\theta = \dfrac{{(2n + 1)\pi }}{2} and θ=nπ+(−1)n(7π6)\theta = n\pi + {( - 1)^n}(\dfrac{{7\pi }}{6}).

Note: The solution of sin⁡θ=−12\sin \theta = \dfrac{{ - 1}}{2}should be calculated carefully. The sign of sin is negative which means you should consider sin in either the 3 rd or 4 th quadrant. Also, the value of cos⁡θ\cos \theta should be calculated carefully as the sign is neither positive or negative but straight equal to 0.