Solve.  `sin^(2)x-3cos^(2)x=sin(2x)`  Help me please.

Expert Answers
lemjay eNotes educator| Certified Educator


To solve, apply the double angle identity of sine which is `sin (2A)=2sinAcosA` .

`sin^2 x-3cos^2x=2sinxcosx`

Then, one side of the function equal to zero and factor the other side.



Next, set each factor equal to zero and solve for x.

For the first factor,

`sinx + cosx= 0`


Divide both sides by cos x, to express it with one trigonometric function.


`tanx =-1`

Then, take the arctangent of the right side.

`x= tan^(-1) (-1)`

`x= 135^o`

Since there is no indicated interval for x, use the general solution of tangent which is

`theta = tan^(-1) y + 180n`

where n is any integer.

So, one of the solutions is:

`x= 135+180n` degrees

And for the second factor,



Then, express the equation with one trigonometric function only.





`x=71.57+180n` degrees

Hence, the solutions to the given trigonometric equation are:

`x_1= 71.57+180n` degrees


`x_2= 135+180n` degrees.


oldnick | Student





`(4tan^2x)/(1+tan^2x) -3= (2tanx)/(1+tan^2x)`




`tan^2x-2tanx +1-3=1`





`tanx=-1`      `rArr`   `x=3/4 pi + k pi`

`tanx=3`         `rArr`    `x=71^0 33' 54''+k pi`