Please solve q18 ?

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Answer 1 · 2018-05-02T18:36:35

$\to {cos}^{6} x + 8 {cos}^{2} x = 4 + 4 {cos}^{4} x$ $rarrcos^6x+8cos^2x=4+4cos^4x$

$\to sin x + {sin}^{2} x + {sin}^{3} x = 1$ $rarrsinx+sin^2x+sin^3x=1$

$\to sin x (1 + {sin}^{2} x) = 1 - {sin}^{2} x = {cos}^{2} x$ $rarrsinx(1+sin^2x)=1-sin^2x=cos^2x$

$\to {sin}^{2} x (1 + 2 {sin}^{2} x + {sin}^{4} x) = {cos}^{4} x$ $rarrsin^2x(1+2sin^2x+sin^4x)=cos^4x$

$\to {sin}^{2} x (1 + 2 {sin}^{2} x + {(1 - {cos}^{2} x)}^{2}) = {cos}^{4} x$ $rarrsin^2x(1+2sin^2x+(1-cos^2x)^2)=cos^4x$

$\to {sin}^{2} x (1 + 2 (1 - {cos}^{2} x) + 1 - 2 {cos}^{2} x + {cos}^{4} x) = {cos}^{4} x$ $rarrsin^2x(1+2(1-cos^2x)+1-2cos^2x+cos^4x)=cos^4x$

$\to {sin}^{2} x (1 + 2 - 2 {cos}^{2} x + 1 - 2 {cos}^{2} x + {cos}^{4} x) = {cos}^{4} x$ $rarrsin^2x(1+2-2cos^2x+1-2cos^2x+cos^4x)=cos^4x$

$\to (1 - {cos}^{2} x) (4 - 4 {cos}^{2} x + {cos}^{4} x) = {cos}^{4} x$ $rarr(1-cos^2x)(4-4cos^2x+cos^4x)=cos^4x$

$rarr4-4cos^2xcancel(+cos^4x)-4cos^2x+4cos^4x-cos^6x=cancelcos^4x$

$rarrcos^6x+8cos^2x=4+4cos^4x$