Question 990313
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I will help you to solve the equation a)


{{{sin^2(2x)}}} + {{{3sin^2(x)}}} = {{{3}}}.                             (1)


From &nbsp;<B>Trigonometry</B>, &nbsp;there is an identity


{{{sin(2x)}}} = {{{2sin(x)*cos(x)}}}.


Substitute it into your equation. &nbsp;You will get


{{{4sin^2(x)*cos^2(x)}}} + {{{3sin^2(x)}}} = {{{3}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)


Recall that {{{cos^2(x)}}} = {{{1-sin^2(x)}}}. 


Substitute it into your equation (2). &nbsp;You will get


{{{4sin^2(x)*(1-sin^2(x))}}} + {{{3sin^2(x)}}} = {{{3}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)


To solve this equation, introduce new variable &nbsp;{{{y}}} = {{{sin^2(x)}}}. &nbsp;Then you will get the equation 


{{{4y*(1-y)}}} + {{{3y}}} = {{{3}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(4)


instead of (3). &nbsp;Simplify &nbsp;(4):


{{{4y^2 - 7y +3}}} = {{{0}}}.


Solve this quadratic equation. &nbsp;Use the quadratic formula. &nbsp;Its roots are &nbsp;y = 1 &nbsp;and &nbsp;y = {{{3/4}}}. 


So, &nbsp;now you need to solve two equations:


1) {{{sin^2(x)}}} = {{{1}}}, &nbsp;or &nbsp;{{{sin(x)}}} = +/- {{{1}}}.


and


2) {{{sin^2(x)}}} = {{{3/4}}}, &nbsp;or &nbsp;{{{sin(x)}}} = +/- {{{sqrt(3)/2}}}.


Please do it yourself.