Solving 2 − 2 sin 2 ⁡ ( x ) − 3 2 sin ⁡ ( 2 x ) = 0I need to solve this equation(preparation for exam). 2 − 2 sin 2 ⁡ ( x ) − 3 2 sin ⁡ ( 2 x ) = 0I have a idea to convert all sin function to "x", and the solve it using quadratic equation, but I am not sure how to do it, because, here I have two different sin function sin 2 ⁡ ( x ) and sin(2x).

Question

Solving   2  −  2      sin    2    ⁡  (  x  )  −      3    2    sin  ⁡  (  2  x  )  =  0I need to solve this equation(preparation for exam).  2  −  2      sin    2    ⁡  (  x  )  −      3    2    sin  ⁡  (  2  x  )  =  0I have a idea to convert all sin function to &quot;x&quot;, and the solve it using quadratic equation, but I am not sure how to do it, because, here I have two different sin function       sin    2    ⁡  (  x  ) and sin(2x).

Carly Yang · Accepted Answer

Step 1You can make the substitution       sin    2    ⁡      x    =  (  1  −      cos    2    ⁡      x    ) and   sin  ⁡      2    x    =  2  sin  ⁡      x    cos  ⁡      x   therefore simplifying the equation down to:  2  −  2  (  1  −      cos    2    ⁡      x    )  −      3    2    (  2  sin  ⁡      x    cos  ⁡      x    )  =  0further simplifying to:  2      cos    2    ⁡      x    −  3  sin  ⁡      x    cos  ⁡      x    =  0which you can then factorise to obtain the solution like you would with an equation of the form   x  (  x  −  a  )  =  0 that gives solution   x  =  0 and   x  =  a:  cos  ⁡      x      (  2  cos  ⁡      x    −  3  sin  ⁡      x    )  =  0Since we know that a solution to the above factored expression is when   cos  ⁡      x    =  0 since if substituted back into the expression it would be   0  ×  (  2  cos  ⁡      x    −  3  sin  ⁡      x    ) and obviously anything multiplied by zero gives zero, so we know that   cos  ⁡      x    =  0 is a solution. We can do this similarly to   (  2  cos  ⁡      x    −  3  sin  ⁡      x    ) since whenever this is zero the expression is also true since it would be   cos  ⁡      x    ×  0 which is obviously zero and therefore satisfies out expression.Therefore we proceed by solving the two equations:  cos  ⁡      x    =  0      and      (  2  cos  ⁡      x    −  3  sin  ⁡      x    )  =  0Step 2Normally when you get these questions there is a restriction on the value of x (e.g.   0  &amp;lt;  x  &amp;lt;  2  π) since by the nature of trigonometric functions they tend to repeat themselves and thus there are often infinite solutions for x. But if there isn&#039;t a restriction one solve it as follows:   cos  ⁡      x    =  0 gives solutions  x  =      π    2    +  n  π      where   n   is an integerlikewise for  (  2  cos  ⁡      x    −  3  sin  ⁡      x    )  =  0  →  tan  ⁡      x    =      2    3  we attain solutions:  x  =      tan          −      1        ⁡            2      3        +  n  πso the general answer is:  x  =      π    2    +  n  π   and   x  =      tan          −      1        ⁡            2      3        +  n  πif the question did indeed say that x was restricted all you would have to do is put in integer values of n like -1,1,2,... into the above solutions and find those values of x that satisfy the restriction (like for the example those x that are less than   2  π but are greater than 0)

Solving 2-2 sin^2(x)-3/2 sin(2x)=0

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New Questions in Algebra I