# 7sin2 θ + 3 cos2θ = 4 show that tan θ = 1/√3pls.. i hv a test tmrw... if u cud answer URGENTLY!!

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### 2 Answers

`sin2theta `

`= 2sintheta*costheta`

`= (2sintheta*costheta)/1`

`= (2sintheta*costheta)/(sin^2theta+cos^2theta)`

`= (2sintheta*costheta)/(cos^2theta((sin^2theta)/(cos^2theta)+sin^2theta))`

`= (2sintheta/costheta)/((sin^2theta/cos^2theta)+1)`

`= (2tantheta)/(1+tan^2theta)`

In the same manner you can derive a term for `cos2theta` using tantheta.

`cos2theta`

`= cos^2theta-sin^2theta`

`= (cos^2theta-sin^2theta)/1`

`= (cos^2theta-sin^2theta)/(cos^2theta+sin^2theta)`

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`= (1-tan^2theta)/(1+tan^2theta)`

`7sin2theta + 3 cos2theta = 4`

`(7*2tantheta)/(1+tan^2theta)+(3(1-tan^2theta))/(1+tan^2theta)= 4`

`14tantheta+3-3tan^2theta = 4(1+tan^2theta)`

`14tantheta+3-3tan^2theta = 4+4tan^2theta`

`7tan^2theta-14tantheta+1 = 0`

If `tantheta = t`

`7t^2-14t+1 = 0`

The solution for this type of quadratic equation is given by;

`t = (14+-sqrt(14^2-4*1*7))/(2*7)`

t = 1.93 OR t = 0.0742

Therefore;

`tantheta = 1.93`

`tantheta = 0.0742`

*The value given in the answer is wrong.*

*Justification for error is given below.*

If `tantheta = 1/sqrt3 then theta = 30`

`7sin2theta + 3 cos2theta = 4`

LHS

`7sin60 + 3cos60`

`= 7*sqrt3/2+3*1/2`

`= (7sqrt3+1)/2`

RHS = 4

`LHS != RHS `

**Sources:**

In the same manner you can derive a term for using tantheta.

.

.

.

If

The solution for this type of quadratic equation is given by;

t = 1.93 OR t = 0.0742

Therefore;

The value given in the answer is wrong.

Justification for error is given below.

If

LHS

RHS = 4

Thanx a ton for ur help :)