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Question: How do you solve \(\sin 2x-\cos x-2\sin x+1=0\)?...

How do you solve sin⁡2x−cos⁡x−2sin⁡x+1=0\sin 2x-\cos x-2\sin x+1=0?

Explanation

Solution

In the question we have the multiple angle which is sin⁡2x\sin 2x. So, we will apply the known formula sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x in the given equation. Now we will take cos⁡x\cos x as common from the terms 2sin⁡xcos⁡x2\sin x\cos x, cos⁡x\cos x and simplify the obtained equation. Again, we will observe the terms in the obtained equation and simplify the equation and equate the terms in the equation to the zero. Now we will get two equations, from these equations we will solve each one by considering the trigonometric values of different ratios. Now we will get the required solution.

Complete step-by-step answer:
Given that sin⁡2x−cos⁡x−2sin⁡x+1=0\sin 2x-\cos x-2\sin x+1=0.
We have the trigonometric formula sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x. Substituting this value in the given equation, then we will get
sin⁡2x−cos⁡x−2sin⁡x+1=0 ⇒2sin⁡xcos⁡x−cos⁡x−2sin⁡x+1=0 \begin{aligned} & \sin 2x-\cos x-2\sin x+1=0 \\\ & \Rightarrow 2\sin x\cos x-\cos x-2\sin x+1=0 \\\ \end{aligned}
Taking cos⁡x\cos x common from the terms 2sin⁡xcos⁡x2\sin x\cos x, cos⁡x\cos x in the above equation, then we will get
⇒cos⁡x(2sin⁡x−1)−2sin⁡x+1=0\Rightarrow \cos x\left( 2\sin x-1 \right)-2\sin x+1=0
Taking −1-1 common from the terms 2sin⁡x2\sin x, 11 in the above equation, then we will get
⇒cos⁡x(2sin⁡x−1)−1(2sin⁡x−1)=0\Rightarrow \cos x\left( 2\sin x-1 \right)-1\left( 2\sin x-1 \right)=0
Again taking 2sin⁡x−12\sin x-1 common from the above equation, then we will get
⇒(2sin⁡x−1)(cos⁡x−1)=0\Rightarrow \left( 2\sin x-1 \right)\left( \cos x-1 \right)=0
Equating each term individually to the zero, then we will have
2sin⁡x−1=02\sin x-1=0 or cos⁡x−1=0\cos x-1=0.
Simplifying the above equations then we will get
2sin⁡x−1=0 ⇒2sin⁡x=1 ⇒sin⁡x=12 \begin{aligned} & 2\sin x-1=0 \\\ & \Rightarrow 2\sin x=1 \\\ & \Rightarrow \sin x=\dfrac{1}{2} \\\ \end{aligned} or cos⁡x−1=0 ⇒cos⁡x=1 \begin{aligned} & \cos x-1=0 \\\ & \Rightarrow \cos x=1 \\\ \end{aligned}
We have sin⁡(π6)=12\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2} and cos⁡π=1\cos \pi =1, then the solutions of above equations are
x=nπ+(−)nπ6x=n\pi +{{\left( - \right)}^{n}}\dfrac{\pi }{6} or x=2kπx=2k\pi where n,k∈Zn,k\in Z.

Note: We can also plot a graph of the given equation with is given below

From this graph also we can find the solution of the given equation by observing the points where the given equation meets the x−axisx-axis. The points where the given equation meets the x−axisx-axis are the solutions for the given equation.