Can someone help? Difficult hard math equation2^(sin3x/sinx)-5*2^cos2x+2=0?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember the formula of sine of triple angle such that:

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

You need to substitute `3sin x - 4sin^3 x`  for `sin 3x`  in `(sin3x)/sinx`  such that:

`(sin3x)/sinx = (3sin x - 4sin^3 x)/sinx`

`(sin3x)/sinx = (3sin x)/sin x - (4sin^3 x)/sinx`

`(sin3x)/sinx = 3 - 4sin^2 x`

Hence `2^((sin3x)/sinx) = 2^(3 - 4sin^2 x) = 2^3*2^(-4sin^2 x).`

You need to substitute `2^3*2^(-4sin^2 x)`  for `2^((sin3x)/sinx)`  in equation such that:

`8*2^(-4sin^2 x) - 5*2^(cos2x) + 2 = 0`

You need to remember the formula of cosine of double angle such that:

`cos 2x = 1 - 2sin^2 x`

`8*2^(-4sin^2 x) - 5*2^(1 - 2sin^2 x) + 2 = 0`

`8*2^(-4sin^2 x) - 10*2^(- 2sin^2 x) + 2 = 0`

Using the following identity `a^(-x) = 1/(a^x)`  yields:

`8/2^(4sin^2 x) - 10/2^(2sin^2 x) + 2 = 0`

You should come up with the following substitution such that:

`2^(sin^2 x) = y => 2^(2sin^2 x) = y^2 and 2^(4sin^2 x) = y^4`

`8/(y^4) - 10/(y^2) + 2 = 0`

`8 - 10y^2 + 2y^4 = 0`

Reducing by 2 yields:

`y^4 - 5y^2 + 4 = 0`

You should use the next substitution such that: `y^2 = t`

`t^2 - 5t + 4 = 0`

Using quadratic formula yields:

`t_(1,2) = (5+-sqrt(25 - 16))/2`

`t_(1,2) = (5+-sqrt9)/2`

`t_(1,2) = (5+-3)/2 => t_1 = 4 ; t_2 = 1`

You should solve the following equations such that:

`y^2 = 4 => y_(1,2) = +-2`

`y^2 =1 => y_(3,4) = +-1`

You should solve for x the following equations such that:

`2^(sin^2 x) = y_(1,2,3,4)`

`2^(sin^2 x) = 2 => sin^2 x = 1 => sin x = +-1`

`x = (-1)^n*(pi/2) + npi`

`x = (-1)^(n+1)*(pi/2) + npi`

`2^(sin^2 x) = -2`  invalid equation since `2^(sin^2 x)`  cannot be negative

`2^(sin^2 x) = 1 => sin^2 x = 0 => sin x = 0 => x = npi`

`x = (-1)^n*pi + npi ; x = x = (-1)^n*2pi + npi`

`2^(sin^2 x) = -1`  invalid equation since `2^(sin^2 x)`  cannot be negative

Hence, evaluating the general solutions to the given equation yields `x = (-1)^n*(pi/2) + npi ; x = (-1)^(n+1)*(pi/2) + npi ; x = npi ; x = (-1)^n*pi + npi ; x = x = (-1)^n*2pi + npi.`

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