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Question Number 83654 by jagoll last updated on 05/Mar/20

$$\mathrm{solve}\mathrm{this}\mathrm{equation}$$$$\sin {}^{2}x-\sin {}^{4}x=\cos {}^{2}x-\cos {}^{4}x$$

Commented by jagoll last updated on 05/Mar/20

$$\iff \cos {}^{4}x-\sin {}^{4}x=\cos {}^{2}x-\sin {}^{2}x$$$$\Rightarrow \cos {}^{2}x-\sin {}^{2}x=\cos {}^{2}x-\sin {}^{2}x$$$$\therefore \mathrm{always}\mathrm{true}\mathrm{for}x\in \mathbb{R}$$$$$$

Answered by MJS last updated on 05/Mar/20

$${\cos }^{2}x=1-{\sin }^{2}x$$$${\sin }^{2}x-{\sin }^{4}x=1-{\sin }^{2}x-{(1-{\sin }^{2}x)}^2$$$${\sin }^{2}x-{\sin }^{4}x=(1-{\sin }^{2}x){\sin }^{2}x$$$$\mathrm{lhs}=\mathrm{rhs}$$$$\mathrm{true}\mathrm{\forall }x\in \mathbb{R}$$

Commented by jagoll last updated on 05/Mar/20

$$\mathrm{yess}...$$

Answered by $@ty@m123 last updated on 05/Mar/20

$$\sin {}^{2}x(1-\sin {}^{2}x)=\cos {}^{2}x(1-\cos {}^{2}x)$$$$\sin {}^{2}x\cos {}^{2}x=\cos {}^{2}x\sin {}^{2}x$$

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