how do you solve 2+cos2x=3sin2x for x?

Question

please show all steps to solve the identity

Answer 1

2+cos2x=3sin2x

Since:
cos²2x + sin²2x =1 then:
sin2x =√(1 - cos²2x)

2+cos2x=3 * √(1 - cos²2x)

Raising to 2:
(2+cos2x) ² = [ 3 * √(1 - cos²2x) ]²
4 + 4cos2x + cos²2x = 9 (1 - cos²2x)
4 + 4cos2x + cos²2x = 9 - 9cos²2x)
10cos²2x + 4cos2x - 5 = 0

IF t = cos2x then:
10t² + 4t - 5 = 0
Solving the quadratic equation we find:

t1 = -0.934846922834953 = cos 2x
t2 = 0.534846922834954 = cos 2x

First we must test both solutions in:
2+cos2x=3 * √(1 - cos²2x)

For t1 = -0.9348469228 = cos 2x:
2-0.9348469228 =3 * √(1 -( -0.9348469228) ²)
OK, it verify equation.

For t2 = 0.5348469228 = cos 2x:
2+ 0.5348469228 =3 * √(1 - 0.5348469228 ²)
OK, it verify equation.

So:
1st:
Cos2x = -0.9348469228
2x = arccos (-0.9348469228)
2x = 159.2034283° + 360° * K
x = 79.60171415° + 180° * K

2nd:
Cos2x = 0.5348469228
2x = arccos 0.5348469228 )
2x = 57.66646931° + 360° * K
x = 28.83323466° + 180° * K

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Solution:
x1 = 79.6017141696649° + 180° * K
x2 = 28.8332346532571° + 180° * K
where K is an integer value.

Answer 2

Let x = 0
Then we have 2+1 =3* 0 --> 3=0
Therefore this is not an identity.

Answer 3

problem a million: Use the quadratic formulation. x = [ -b ± ?(b² - 4ac) ] / 2a a = a million b = a million c = -3 x = [ -a million ± ?(a million² - 4(a million)(-3)) ] / 2 x = [ -a million ± ?(13) ] / 2 answer: x = -a million/2 ± (?13)/2 problem A: Quadratic formulation lower back with a = 5, b = -4, c = a million. you attempt this one on your guy or woman... problem B: 9x² = 14 x² = 14/9 x = ± ?(14/9) x = ± (?14) / 3 problem C: Take the sq. root of the two facets: x + 5 = ± ?3 x = -5 ± ?3 problem D; Quadratic formulation problem E: See problem a million.

Answer 4

2 + cos2x = 3sin2x
2 + (cos^2 x - sin^2 x) = 3(2sinx cosx)
2 + (1 - sin^2 x - sin^2 x) = 6(sinx cosx)
3 - 2sin^2 x = 6(sinx cosx)
3(1) - 2sin^2 x = 6(sinx cosx)
since 1 = sin^2 x + cos^2 x
3(sin^2 x + cos^2 x) - 2sin^2 x = 6(sinx cosx)
rearranging
3(sin^2 x + cos ^2 x) - 6(sinx cosx ) =2sin^2 x
3(sin^2 x + cos^2 x - 2sinx cosx) = 2sin^2 x
3(sinx - cosx)^2 = 2sin^2 x
(sinx -cosx)^2 = 2/3(sin^2 x)
sinx - cosx = + or - sqrt(2/3)(sinx)
dividing by sinx
1 - cot x = + or - sqrt(2/3)
cot x = 1 + sqrt(2/3) or 1 - sqrt(2/3)

x = cot inverse(1+ sqrt(2/3)) or cot inverse (1 - sqrt(2/3))

Answer 5

1) 2+cos(2x)=3sin(2x),

[2+cos(2x)]^2=[3sin(2x)]^2 ,

4+ 4·cos(2x)+ cos^2(2x) = 9 sin^2(x)

4+ 4·cos(2x)+ cos^2(2x) = 9 [1-cos^2(x)]

4+ 4·cos(2x)+ cos^2(2x) = 9 [1-cos^2(x)]

10· cos^2(2x) + 4·cos(2x) - 5 = 0

cos(2x) = [-2 ±√(4+50)]/10 =[-2 ±3√6]/10

*cos(2x) = [-2 +3√6]/10
--> x=28º50' +180º·k or
--> ¡ the other isn't solution x= 151º10' + 180º·k!

*cos(2x)=[-2-3√6]/10
--> x=79º36'6'' +180º·k

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2) 2+cos^2x=3sin^2x

2+ 1 -sin^2x =3·sin^2x

4·sin^2x=3
sin^2x=3/4

sinx = ±√3 / 2 -->

x=60º+ 360º · k
x=120º+360º·k
x=240º+360º·k
x=300+360º·k with k any integer.

Saludos.