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Question Number 15471 by Mr easymsn last updated on 10/Jun/17

Answered by sma3l2996 last updated on 10/Jun/17

$$\left(a\right)$$$$2sin\left(x\right)cos\left(x\right)+sinx=1-2{cos}^{2}x-cosx$$$$sin\left(x\right)(2cos\left(x\right)+1)=1-2{cos}^{2}x-cosx$$$$letcosx=t$$$$\sqrt{1-t^2}(2t+1)=1-2t^2-t$$$$(1-t^2){(2t+1)}^2={(1-2t^2-t)}^2$$$$(1-t^2)(4t^2+4t+1)={(t+1)}^2{(-2t+1)}^2$$$$(1-t)(1+t)(4t^2+4t+1)={(t+1)}^2(4t^2-4t+1)$$$$(1-t)(4t^2+4t+1)=(t+1)(4t^2-4t+1)$$$$4t^2+4t+1-4t^3-4t^2-t=4t^3-4t^2+t+4t^2-4t+1$$$$-4t^3+3t=4t^3-3t$$$$t(8t^2-6)=0\iff t=0or{t}^2=\frac{3}{4}$$$$t=0ort=\frac{\underset{-}{+}\sqrt{3}}{2}$$$$x=\frac{\underset{-}{+}\pi }{2},\underset{-}{+}\frac{\pi }{6},\underset{-}{+}\frac{5\pi }{6}$$$$\left(b\right)$$$$2{sin}^{2}x+{sin}^2\left(2x\right)-2=0$$$$2{sin}^{2}x+4{sin}^2\left(x\right){cos}^2\left(x\right)-2=0$$$${sin}^{2}x+2{sin}^{2}x(1-{sin}^{2}x)-1=0$$$${sin}^{2}x+2{sin}^{2}x-2{sin}^{4}x-1=0$$$$2{sin}^{4}x-3{sin}^{2}x+1=0\iff ({sin}^{2}x-1)(2{sin}^{2}x-1)=0$$$${sin}^{2}x=1or{sin}^{2}x=\frac{1}{2}$$$$sinx=\underset{-}{+}1orsinx=\underset{-}{+}\frac{\sqrt{2}}{2}$$$$x=\underset{-}{+}\frac{\pi }{2},\underset{-}{+}\frac{\pi }{4},\underset{-}{+}\frac{3\pi }{4}$$$$\left(c\right)$$$$3sinx+\sqrt{3cosx}=3$$$$\sqrt{3cosx}=3(1-sinx)$$$$3cosx=9(1+{sin}^{2}x-2sinx)=9(2-{cos}^{2}x-2sinx)$$$$cosx=6-3{cos}^{2}x-6sinx$$$$3{cos}^{2}x+cosx-6=-6sinx$$$$lett=cosx$$$$3t^2+t-6=-6\sqrt{1-t^2}$$$${(3t^2+t-6)}^2=36-36t^2$$$$9t^4+6t^2(t-6)+t^2-12t+36=36-36t^2$$$$9t^4+6t^3-36t^2+t^2-12t=-36t^2$$$$9t^4+6t^3+t^2-12t=0$$$$t(9t^3+6t^2+t-12)=0$$$$$$

Commented by myintkhaing last updated on 11/Jun/17

$$cos2x=2{cos}^{2}x-1$$

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