Solve `cosx+sinx=2/3` ?

To solve this equation, we have to use trigonometric functions to isolate one of the unknowns and solve for that unknown.

In this case, we will use `sin^2`(`x`) `+` `cos^2`(`x`) = 1 and 2`sin(x)``cos(x)` = `sin(2x)`.

`sin`(`x`) `+` `cos`(`x`) = `2/3`

`(sin(x) + cos(x))^2` `=` `(2/3)^2 =` `4/9`

`sin^2(x)` `+` `cos^2(x)` `+` 2`sin(x)``cos(x)` `=` `4/9`

`1` `+` `sin(2x)` `=` `4/9`

`sin(2x)` `=` `- 5/9`

The function `sin(x)` is negative in the 3rd and 4th quadrants and `sin(theta)=sin(180^"o"-theta)`. Therefore, we have two solutions for `2x`:

`2x_1` `=` `sin^-1(-5/9)` `=` `-33.75^"o"`
`2x_2 = 180^"o"-(-33.75^"o") = 213.75^"o"`

From this:
`x_1` `=` `-33.75^"o"/2` `=` `-16.87^"o"`
`x_2 = 213.75^"o"/2 = 106.87^"o"`

`cos(x_1)=cos(-16.87^"o")` `=` `0.96`
`cos(x_2)=cos(106.87^"o")=-0.29`

`sinx+cosx=2/3`

Let's first use linear combination of cosine and sine with equal arguments formula to simplify it.

That is, we want to express
A cos x +B sin x in the form C cos (x-D). Note that A is the coefficient of cos x and B is the coefficient of sine x.

To find C use pythagorean formula and to find D we use one of these two formulas
`cos D = A/C, sinD=B/C`

From the given equation A = 1 and B = 1. So let's find C using pythagorean theorem.
`C=sqrt(A^2+B^2) = sqrt(1^2+1^2)=sqrt 2`

To find D we need to first figure out which quadrant x is in and because both cos x and sin x are positive it means that x is in quadrant one.

`cos D=1/sqrt2`
`D=cos^-1 (1/sqrt2) = pi/4`

Therefore `sinx+cosx=sqrt2 cos (x-pi/4)`

Now let's use that to solve the problem. That is,

`sqrt2 cos (x-pi/4)=2/3`

`cos (x-pi/4)=2/(3sqrt2)`

`x-pi/4=cos^-1(2/(3sqrt2))`

`x=cos^-1(2/(3sqrt2)) + pi/4`

`x=1.87`-Round to 2 decimal places

Making `y = sin(x)` and `alpha = 2/3`

`y+sqrt(1-y^2) = alpha`
`1-y^2=(alpha-y)^2`
`1-y^2=alpha^2-2alpha y+y^2`
`2y^2-2alpha y+alpha^2-1=0` solving for `y`

`y= 1/2(alpha pm sqrt(2-alpha^2))`

`sin(x) = 1/2 (2/3 pm sqrt[14]/3)`

Considering the given condition is

`cosx+sinx=2/3......(1)`

Now

`(cosx-sinx)^2+(cosx+sinx)^2=2(cos^2x+sin^2x)=2`

`=>(cosx-sinx)^2+(2/3)^2=2`

`=>(cosx-sinx)^2+(2/3)^2=2`

`=>(cosx-sinx)^2=2-4/9=14/9`

`=>(cosx-sinx)=+-sqrt14/3.....(2)`

Adding (1) and (2) we get

`2cosx=2/3+-sqrt14/3`

`=>cosx=1/2(2/3+-sqrt14/3)`