3cos^xs-2sin2x+sin^x=1

$3{\cos}^2x-2\sin 2x+{\sin}^2x=1$ $\Rightarrow3{\cos}^2x-2\sin 2x+{\sin}^2x={\sin}^2x+{\cos}^2x$ $\Rightarrow 2{\cos}^2x-2\sin 2x=0$ $\Rightarrow 2{\cos}^2x-4\sin x\cos x=0$ $\Rightarrow 2\cos x(\cos x-2\sin x)=0$ $\Rightarrow \left[ \begin{array}{l} \cos x=0(1) \\ \cos x-2\sin x=0 (2)\end{array} \right .$ (1) $\Rightarrow x=\dfrac{\pi}{2}+k\pi(k\in\mathbb Z)$ (2) $\Rightarrow \dfrac{1}{\sqrt5}\cos x-\dfrac{2}{\sqrt5}\sin x=0$ Đặt $\cos \alpha=\dfrac{1}{\sqrt5}$; $\sin \alpha=\dfrac{2}{\sqrt5}$ Phương trình tương đương $\cos\alpha\cos x-\sin\alpha\sin x=0$ $\Rightarrow \cos(x+\alpha)=0$ $\Rightarrow x+\alpha=\dfrac{\pi}{2}+k\pi$ $\Rightarrow x=-\alpha+\dfrac{\pi}{2}+k\pi(k\in\mathbb Z)$

3cos2x−2sin2x+sin2x=13cos2x−2sin⁡2x+sin2x=1 ⇒3cos2x−2sin2x+sin2x=sin2x+cos2x⇒3cos2x−2sin⁡2x+sin2x=sin2x+cos2x ⇒2cos2x−2sin2x=0⇒2cos2x−2sin⁡2x=0 ⇒2cos2x−4sinxcosx=0⇒2cos2x−4sin⁡xcos⁡x=0 ⇒2cosx(cosx−2sinx)=0⇒2cos⁡x(cos⁡x−2sin⁡x)=0 ⇒[cosx=0(1)cosx−2sinx=0(2)⇒[cos⁡x=0(1)cos⁡x−2sin⁡x=0(2) (1) ⇒x=π2+kπ(k∈Z)⇒x=π2+kπ(k∈Z) (2) ⇒1√5cosx−2√5sinx=0⇒15cos⁡x−25sin⁡x=0 Đặt cosα=1√5cos⁡α=15; sinα=2√5sin⁡α=25 Phương trình tương đương cosαcosx−sinαsinx=0cos⁡αcos⁡x−sin⁡αsin⁡x=0 ⇒cos(x+α)=0⇒cos⁡(x+α)=0 ⇒x+α=π2+kπ⇒x+α=π2+kπ ⇒x=−α+π2+kπ(k∈Z)

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