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Question: How do you solve \( \sin \left( 2x \right)=-\cos \left( 2x \right) \) ?...

How do you solve sin⁡(2x)=−cos⁡(2x)\sin \left( 2x \right)=-\cos \left( 2x \right) ?

Explanation

Solution

Hint : We first simplify the equation to convert it into the form of ratio tan. Then we find the principal value of x for which tan⁡(2x)=−1\tan \left( 2x \right)=-1 . In that domain, equal value of the same ratio gives equal angles. We find the angle value for x. At the end we also find the general solution for the equation tan⁡(2x)=−1\tan \left( 2x \right)=-1 .

Complete step by step solution:
The given equation is sin⁡(2x)=−cos⁡(2x)\sin \left( 2x \right)=-\cos \left( 2x \right) . We take all the ratios on one side to get
sin⁡(2x)=−cos⁡(2x) ⇒sin⁡(2x)cos⁡(2x)=−1 ⇒tan⁡(2x)=−1 \begin{aligned} & \sin \left( 2x \right)=-\cos \left( 2x \right) \\\ & \Rightarrow \dfrac{\sin \left( 2x \right)}{\cos \left( 2x \right)}=-1 \\\ & \Rightarrow \tan \left( 2x \right)=-1 \\\ \end{aligned}
It’s given that tan⁡(2x)=−1\tan \left( 2x \right)=-1 . The value in is −1-1 . We need to find x for which tan⁡(2x)=−1\tan \left( 2x \right)=-1 .
We know that in the principal domain or the periodic value of −π2≤x≤π2-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} for tan⁡x\tan x , if we get tan⁡a=tan⁡b\tan a=\tan b where −π2≤a,b≤π2-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2} then a=ba=b .
We have the value of tan⁡(−π4)\tan \left( -\dfrac{\pi }{4} \right) as −1-1 . −π2<−π4<π2-\dfrac{\pi }{2}<-\dfrac{\pi }{4}<\dfrac{\pi }{2} .
Therefore, tan⁡(2x)=−1=tan⁡(−π4)\tan \left( 2x \right)=-1=\tan \left( -\dfrac{\pi }{4} \right) which gives 2x=−π42x=-\dfrac{\pi }{4} .
For tan⁡(2x)=−1\tan \left( 2x \right)=-1 , the value of x is x=−π8x=-\dfrac{\pi }{8} .
We also can show the solutions (primary and general) of the equation tan⁡(2x)=−1\tan \left( 2x \right)=-1 through the graph. We take y=tan⁡(2x)=−1y=\tan \left( 2x \right)=-1 . We got two equations y=tan⁡(2x)y=\tan \left( 2x \right) and y=−1y=-1 . We place them on the graph and find the solutions as their intersecting points.

We can see the primary solution in the interval −π2≤x≤π2-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} is the point A as x=−π8x=-\dfrac{\pi }{8} .
All the other intersecting points of the curve and the line are general solutions.

Note : Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to −∞≤x≤∞-\infty \le x\le \infty . In that case we have to use the formula x=nπ+ax=n\pi +a for tan⁡a=tan⁡b\tan a=\tan b where −π2≤a≤π2-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2} . For our given problem tan⁡(2x)=−1\tan \left( 2x \right)=-1 , the general solution will be 2x=nπ−π42x=n\pi -\dfrac{\pi }{4} . Here n∈Zn\in \mathbb{Z} . The simplified form is x=nπ2−π8x=n\dfrac{\pi }{2}-\dfrac{\pi }{8} .