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Question Number 75763 by mathocean1 last updated on 16/Dec/19

please help me to solve it in R 3cosx−(√3)sinx+(√6)=0

$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\mathbb{R} \\ $$$$\mathrm{3cosx}−\sqrt{\mathrm{3}}\mathrm{sinx}+\sqrt{\mathrm{6}}=\mathrm{0} \\ $$

Answered by Ajao yinka last updated on 16/Dec/19

Answered by $@ty@m123 last updated on 17/Dec/19

Divide by 2(√3), ((√3)/2)cos x−(1/2)sin x+(1/(√2))=0 sin (π/3)cos x−cos (π/3)sin x=−(1/(√2)) sin ((π/3)−x)=−(1/(√2)) sin (x−(π/3))=(1/(√2)) sin (x−(π/3))=sin (π/4) x−(π/3)=nπ+(−1)^n (π/4) , n∈Z x=nπ+(−1)^n (π/4)+(π/3)

$${Divide}\:{by}\:\mathrm{2}\sqrt{\mathrm{3}}, \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}=\mathrm{0} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:{x}−\mathrm{cos}\:\frac{\pi}{\mathrm{3}}\mathrm{sin}\:{x}=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−{x}\right)=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)=\mathrm{sin}\:\frac{\pi}{\mathrm{4}} \\ $$$${x}−\frac{\pi}{\mathrm{3}}={n}\pi+\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{4}}\:\:,\:{n}\in\mathbb{Z} \\ $$$${x}={n}\pi+\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{3}} \\ $$

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