Solving \(\displaystyle{\sin{{\left({3}{x}+{28}^{{\circ}}\right)}}}={\cos{{\left({2}{x}-{13}^{{\circ}}\right)}}}\) Solution: 3x+28+2x-13=90 x=15 3x+28-2x+13=90 x=49 But the value x=87 also satisfies the equation.

justbethlflql

justbethlflql

Answered question

2022-04-02

Solving sin(3x+28)=cos(2x13)
Solution:
3x+28+2x-13=90
x=15
3x+28-2x+13=90
x=49
But the value
x=87
also satisfies the equation. I do not know how it comes.

Answer & Explanation

Boehm98wy

Boehm98wy

Beginner2022-04-03Added 18 answers

Note: Using cos(x)=sin(90x) (since you are working in degrees) your equation becomes:
sin(3x+28)=sin(902x+13),x[0,360]
Now, if sin(a)=sin(b) for a,b[0,360] you must have a=b+360k  or  a+b=180k

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