Question-116459

Question Number 116459 by nguyenthanh last updated on 04/Oct/20

Answered by bemath last updated on 04/Oct/20

((1−cos ((π/(12)))sin x)/(cos x)) +((√6) −(√2) )sin (x−((7π)/(12))) = cos ((π/2)+(π/(12))) ((1−cos ((π/(12)))sin x)/(cos x)) + ((√6)−(√2))sin (x−((7π)/(12)))=−sin ((π/(12))) 1+((√6)−(√2))cos x.sin (x−((7π)/(12)))=cos ((π/(12)))sin x−cos x.sin ((π/(12))) 1+(((√6)−(√2))/2).(2cos x sin (π−(x−((7π)/(12))))=sin (x−(π/(12))) 1+(((√6)−(√2))/2).(2sin (((5π)/(12))−x).cos x)=sin (x−(π/(12))) 1+(((√6)−(√2))/2) (sin (((5π)/(12)))+sin (((5π)/(12))−2x))=sin (x−(π/(12))) 1+((((√6)−(√2))/2))((((√6)+(√2))/4))+((((√6)−(√2))/2))sin (((5π)/(12))−2x)=sin (x−(π/(12))) (3/2) −2((((√2)−(√6))/4))sin (((5π)/(12))−2x)=sin (x−(π/(12))) (3/2)+2sin ((π/(12)))sin (((5π)/(12))−2x)=sin (x−(π/(12))) (3/2)+cos ((π/3)−2x)−cos ((π/2)−2x)=sin (x−(π/(12))) (3/2)+cos ((π/3)−2x)=sin (x−(π/(12)))+sin 2x

$$\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{12}}\right)\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\:+\left(\sqrt{\mathrm{6}}\:−\sqrt{\mathrm{2}}\:\right)\mathrm{sin}\:\left(\mathrm{x}−\frac{\mathrm{7}\pi}{\mathrm{12}}\right)\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{12}}\right)\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\:+\:\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)\mathrm{sin}\:\left(\mathrm{x}−\frac{\mathrm{7}\pi}{\mathrm{12}}\right)=−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{12}}\right) \\ $$$$\mathrm{1}+\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)\mathrm{cos}\:\mathrm{x}.\mathrm{sin}\:\left(\mathrm{x}−\frac{\mathrm{7}\pi}{\mathrm{12}}\right)=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{12}}\right)\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}.\mathrm{sin}\:\left(\frac{\pi}{\mathrm{12}}\right) \\ $$$$\mathrm{1}+\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}.\left(\mathrm{2cos}\:\mathrm{x}\:\mathrm{sin}\:\left(\pi−\left(\mathrm{x}−\frac{\mathrm{7}\pi}{\mathrm{12}}\right)\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right)\right. \\ $$$$\mathrm{1}+\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}.\left(\mathrm{2sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{12}}−\mathrm{x}\right).\mathrm{cos}\:\mathrm{x}\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right) \\ $$$$\mathrm{1}+\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left(\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)+\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{12}}−\mathrm{2x}\right)\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right) \\ $$$$\mathrm{1}+\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\right)+\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{12}}−\mathrm{2x}\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\:−\mathrm{2}\left(\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}}{\mathrm{4}}\right)\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{12}}−\mathrm{2x}\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{12}}\right)\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{12}}−\mathrm{2x}\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{2x}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{2x}\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{2x}\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right)+\mathrm{sin}\:\mathrm{2x} \\ $$$$ \\ $$

Answered by nguyenthanh last updated on 04/Oct/20

$${x}=? \\ $$

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