if-sin-x-cos2x-3-

Question Number 198249 by liuxinnan last updated on 15/Oct/23

$${if}\:\:{sin}\left({x}+\varphi\right)+{cos}\mathrm{2}{x}\leqslant\sqrt{\mathrm{3}\:}\: \\ $$$$\varphi=? \\ $$

Commented by Frix last updated on 16/Oct/23

I get one solution ϕ=(π/4)+(1/2)sin^(−1) (((7(√7)−19)(√3))/9) ⇒ maximum (sin (x+ϕ) +cos (2x)) =(√3) at x=(1/2)cos^(−1) ((((√7)−1)(√3))/3) [x≈.158499811256∧ϕ≈.739169245092]

$$\mathrm{I}\:\mathrm{get}\:\mathrm{one}\:\mathrm{solution} \\ $$$$\varphi=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \:\frac{\left(\mathrm{7}\sqrt{\mathrm{7}}−\mathrm{19}\right)\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$$\Rightarrow\:\mathrm{maximum}\:\left(\mathrm{sin}\:\left({x}+\varphi\right)\:+\mathrm{cos}\:\left(\mathrm{2}{x}\right)\right)\:=\sqrt{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\mathrm{at}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \:\frac{\left(\sqrt{\mathrm{7}}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\left[{x}\approx.\mathrm{158499811256}\wedge\varphi\approx.\mathrm{739169245092}\right] \\ $$

Commented by mr W last updated on 17/Oct/23

$${i}'{m}\:{interested}\:{what}\:{your}\:{method}\:{is}. \\ $$$${similar}\:{to}\:{my}\:{solution}\:{below}\:{or}\: \\ $$$${quite}\:{different}? \\ $$

Commented by Frix last updated on 17/Oct/23

I don′t yet understand why my method works. I solved the given equation for ϕ. You get the function ϕ(x). While inserting different values of x I found min (ϕ(x)) gives the searched value.

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{yet}\:\mathrm{understand}\:\mathrm{why}\:\mathrm{my}\:\mathrm{method} \\ $$$$\mathrm{works}.\:\mathrm{I}\:\mathrm{solved}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{for}\:\varphi. \\ $$$$\mathrm{You}\:\mathrm{get}\:\mathrm{the}\:\mathrm{function}\:\varphi\left({x}\right).\:\mathrm{While}\:\mathrm{inserting} \\ $$$$\mathrm{different}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{I}\:\mathrm{found}\:\mathrm{min}\:\left(\varphi\left({x}\right)\right) \\ $$$$\mathrm{gives}\:\mathrm{the}\:\mathrm{searched}\:\mathrm{value}. \\ $$

Commented by mr W last updated on 17/Oct/23

$${thanks}! \\ $$

Answered by mr W last updated on 16/Oct/23

f(x)=sin (x+ϕ)+cos 2x=(√3) f′(x)=cos (x+ϕ)−2 sin 2x=0 ((√3)−cos 2x)^2 +(2 sin 2x)^2 =1 3−2(√3) cos 2x+cos^2 2x+4 sin^2 2x=1 3−2(√3) cos 2x+3(1−cos^2 2x)=0 3 cos^2 2x+2(√3) cos 2x−6=0 ⇒cos 2x=(((√(21))−(√3))/3) ⇒x=(1/2)cos^(−1) (((√(21))−(√3))/( 3)) cos (x+ϕ)=2×(√(1−((((√(21))−(√3))/( 3)))^2 ))=((2(√(6(√7)−15)))/3) ⇒ϕ=cos^(−1) ((2(√(6(√7)−15)))/3)−(1/2)cos^(−1) (((√(21))−(√3))/( 3)) ≈0.7391692451

$${f}\left({x}\right)=\mathrm{sin}\:\left({x}+\varphi\right)+\mathrm{cos}\:\mathrm{2}{x}=\sqrt{\mathrm{3}} \\ $$$${f}'\left({x}\right)=\mathrm{cos}\:\left({x}+\varphi\right)−\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{x}=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{3}}−\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{2}} +\left(\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}{x}+\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}+\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}=\mathrm{1} \\ $$$$\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}{x}+\mathrm{3}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}\right)=\mathrm{0} \\ $$$$\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}{x}−\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}{x}=\frac{\sqrt{\mathrm{21}}−\sqrt{\mathrm{3}}}{\mathrm{3}}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{21}}−\sqrt{\mathrm{3}}}{\:\mathrm{3}} \\ $$$$\mathrm{cos}\:\left({x}+\varphi\right)=\mathrm{2}×\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{21}}−\sqrt{\mathrm{3}}}{\:\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}\sqrt{\mathrm{6}\sqrt{\mathrm{7}}−\mathrm{15}}}{\mathrm{3}} \\ $$$$\Rightarrow\varphi=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}\sqrt{\mathrm{6}\sqrt{\mathrm{7}}−\mathrm{15}}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{21}}−\sqrt{\mathrm{3}}}{\:\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\approx\mathrm{0}.\mathrm{7391692451} \\ $$

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