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Câu 2$cos^2x-cos^24x-sinx.cos4x=\frac{1}{4}$$<=>(cosx-cos4x)(cosx+cos4x)-sinx.cos4x=\frac{1}{4}$$<=>-2sin\frac{5x}{2}.cos\frac{5x}{2}.2sin\frac{3x}{2}.cos\frac{3x}{2}-\frac{1}{2}sin5x-\frac{1}{2}sin3x-\frac{1}{4}=0$$<=>(sin5x.sin3x+\frac{1}{2}sin5x)+(\frac{1}{2}sin3x+\frac{1}{4})=0 $$<=>sin5x(sin3x+\frac{1}{2})+\frac{1}{2}(sin3x+\frac{1}{2})=0$$<=>(sin3x+\frac{1}{2})(sin5x+\frac{1}{2})=0$Đơn giản rồi bạn tự giải nhé

Câu 2$cos^2x-cos^24x-sinx.cos4x=\frac{1}{4}$$<=>(cosx-cos4x)(cosx+cos4x)-sinx.cos4x=\frac{1}{4}$$<=>2sin\frac{5x}{2}.cos\frac{5x}{2}.2sin\frac{3x}{2}.cos\frac{3x}{2}-\frac{1}{2}sin5x+\frac{1}{2}sin3x-\frac{1}{4}=0$$<=>(sin5x.sin3x-\frac{1}{2}sin5x)+(\frac{1}{2}sin3x-\frac{1}{4})=0 $$<=>sin5x(sin3x-\frac{1}{2})+\frac{1}{2}(sin3x-\frac{1}{2})=0$$<=>(sin3x-\frac{1}{2})(sin5x-\frac{1}{2})=0$Đơn giản rồi bạn tự giải nhé