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((sin(x))/(√(2sin^2 (x)+cos^2 (x)))) +(1/(√2))=csc(x)(√(2sin^2 (x)+cos^2 (x))) show that x={(π/2)+2πn} and x={cos^(−1) ((√3))−π+2πn} and x={−cos^(−1) ((√3))+2πn}

$$\frac{{sin}\left({x}\right)}{\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}={csc}\left({x}\right)\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)} \\ $$$${show}\:{that} \\ $$$${x}=\left\{\frac{\pi}{\mathrm{2}}+\mathrm{2}\pi{n}\right\}\:{and}\:{x}=\left\{{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}\right)−\pi+\mathrm{2}\pi{n}\right\} \\ $$$${and}\:{x}=\left\{−{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}\right)+\mathrm{2}\pi{n}\right\} \\ $$$$ \\ $$

Question Number 84005 Answers: 2 Comments: 4

find atleast 7 solutions of the equation. 900x+7689y=109876 CAN ANYONE SOLVE THIS now lets find 7 integral solutions

$${find}\:{atleast}\:\mathrm{7}\:{solutions} \\ $$$${of}\:{the}\:{equation}. \\ $$$$\mathrm{900}{x}+\mathrm{7689}{y}=\mathrm{109876} \\ $$$${CAN}\:{ANYONE}\:{SOLVE} \\ $$$${THIS} \\ $$$${now}\:{lets}\:{find}\:\mathrm{7}\:{integral} \\ $$$${solutions} \\ $$

Question Number 83941 Answers: 1 Comments: 0

If ((2 ))^(1/3) + (4)^(1/(3 )) + ((8 ))^(1/(3 )) = x then x^3 −6x^2 +6x+6 = ?

$$\mathrm{If}\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{4}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{8}\:}\:=\:\mathrm{x}\: \\ $$$$\mathrm{then}\:\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{6}\:=\:? \\ $$

Question Number 83931 Answers: 1 Comments: 1

(1/(((√1)+(√2))((1)^(1/(4 )) +(2)^(1/(4 )) ))) + (1/(((√2)+(√3))((2)^(1/( 4)) +(3)^(1/(4 )) ))) + (1/(((√3)+(√4))((3)^(1/(4 )) +(4)^(1/(4 )) ))) + ... + (1/(((√(255))+(√(256)))(((255))^(1/(4 )) +((256))^(1/(4 )) ))) = ...

$$\frac{\mathrm{1}}{\left(\sqrt{\mathrm{1}}+\sqrt{\mathrm{2}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{1}}+\sqrt[{\mathrm{4}\:}]{\mathrm{2}}\right)}\:+\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\left(\sqrt[{\:\mathrm{4}}]{\mathrm{2}}+\sqrt[{\mathrm{4}\:}]{\mathrm{3}}\right)}\:+ \\ $$$$\frac{\mathrm{1}}{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{3}}+\sqrt[{\mathrm{4}\:}]{\mathrm{4}}\right)}\:+\:...\:+\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{255}}+\sqrt{\mathrm{256}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{255}}+\sqrt[{\mathrm{4}\:}]{\mathrm{256}}\right)} \\ $$$$=\:...\: \\ $$

Question Number 83910 Answers: 2 Comments: 1

find all 6 digit numbers which are not only palindrome but also divisible by 495.

$$\mathrm{find}\:\mathrm{all}\:\mathrm{6}\:\mathrm{digit}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{only}\:\mathrm{palindrome}\:\mathrm{but}\:\mathrm{also}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{495}. \\ $$

Question Number 83874 Answers: 1 Comments: 1

Question Number 83871 Answers: 0 Comments: 4

If equation { (((√(x^2 +y^2 ))+(√((x−4)^2 +y^2 ))+(√(x^2 +(y−3)^2 ))+(√((x−4)^2 +(y−3)^2 ))=10)),((x+2y= 5z)) :} has solution is (a,b,c). find a+2b+3c

$$\mathrm{If}\:\mathrm{equation}\: \\ $$$$\begin{cases}{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{10}}\\{\mathrm{x}+\mathrm{2y}=\:\mathrm{5z}}\end{cases} \\ $$$$\mathrm{has}\:\mathrm{solution}\:\mathrm{is}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right).\: \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{2b}+\mathrm{3c}\: \\ $$

Question Number 83834 Answers: 1 Comments: 2