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

*print*Print*list*Cite

### 2 Answers

`sin^2x-3cos^2x=sin(2x)`

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.

`sin^2x-2sinxcosx-3cos^2x=0`

`(sinx+cosx)(sinx-3cosx)=0`

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

For the first factor,

`sinx + cosx= 0`

`sinx=-cosx`

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

`sinx/cosx=-cosx/cosx`

`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,

`sinx-3cosx=0`

`sin=3cosx`

Then, express the equation with one trigonometric function only.

`sinx/cosx=(3cosx)/cosx`

`tanx=3`

`x=tan^(-1)3`

`x=71.57^o`

`x=71.57+180n` degrees

**Hence, the solutions to the given trigonometric equation are:**

**`x_1= 71.57+180n` degrees**

**and**

**`x_2= 135+180n` degrees.**

`sin^2x-3cos^2x=sin2x`

`sin^2x-3(1-sin^2x)=2sinxcosx`

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

`4sin^2x-3=2sinxcosx`

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

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

`4tan^2x-3-3tan^2x-2tanx=0`

`tan^2x-2tanx-3=0`

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

`(tanx-1)^2-3=1`

`(tanx-1)^2=4`

`tanx-1=+-2`

`tanx=1+-2`

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

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