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Question Number 159049 by mnjuly1970 last updated on 12/Nov/21

$$$$$$solvethefollowingequation:$$$$$$$${sin}^6\left(x\right)+{cos}^6\left(x\right)+{sin}^4\left(x\right)+{cos}^4\left(x\right)=\frac{3}{4}$$$$$$

Answered by gsk2684 last updated on 12/Nov/21

$$\Rightarrow {(\sin {}^{2}x+\cos {}^{2}x)}^3-3\sin {}^{2}x\cos {}^{2}x(\sin {}^{2}x+\cos {}^{2}x)$$$$+{(\sin {}^{2}x+\cos {}^{2}x)}^2-2\sin {}^{2}x\cos {}^{2}x=\frac{3}{4}$$$$\Rightarrow 1-3\sin {}^{2}x\cos {}^{2}x+1-2\sin {}^{2}x\cos {}^{2}x=\frac{3}{4}$$$$\Rightarrow \frac{5}{4}=5\sin {}^{2}x\cos {}^{2}x$$$$\Rightarrow 1=\sin {}^{2}2x$$$$\therefore 2x=(2n+1)\frac{\pi }{2},n\in Z$$$$x=(2n+1)\frac{\pi }{4},n\in Z$$$$.....gsk....$$

Commented by mnjuly1970 last updated on 12/Nov/21

$$thanksalotsirg.s.k$$

Commented by gsk2684 last updated on 13/Nov/21

$$welcome$$

Answered by MJS_new last updated on 12/Nov/21

$$\sin x=s\wedge \cos x=\sqrt{1-s^2}$$$$\Rightarrow$$$$s^4-s^2+\frac{1}{4}=0$$$${(s^2-\frac{1}{2})}^2=0$$$$\Rightarrow$$$$\sin x=\pm \frac{\sqrt{2}}{2}$$$$\Rightarrow$$$$x=-\frac{\pi }{4}+\frac{n\pi }{2}$$

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