Solve cos(5x-10°)=sin(20-3x) pls help me and pls show the working
Recall that
sin ( 90 - A ) = cos A
cos ( 90 - A ) = sin A
cos(5x-10°)=sin(20-3x)
If A=20-3x
Cos(5x-10)=cos(90-(20-3x)
Cos(5x-10)=cos (90-20+3x)
Cos(5x-10)=cos(70+3x)
Compare
5x-10=70+3x
2x=80
X=40°
To solve the equation cos(5x-10°)=sin(20-3x), we'll start by rewriting the equation using the trigonometric identities.
Recall that sin(x) can be expressed as cos(90° - x). Therefore, we can rewrite sin(20-3x) as cos(90° - (20-3x)).
cos(5x-10°) = cos(90° - (20-3x))
Now, we know that two cosines are equal if their angles are either equal or have a difference of 360° (or 2π radians). So, we can set the angles equal to each other and solve for x.
5x - 10° = 90° - (20 - 3x)
First, let's simplify the right-hand side of the equation:
5x - 10° = 90° - 20 + 3x
Combine like terms:
5x - 3x = 90° + 20 - 10°
2x = 100° - 10°
2x = 90°
Now, solve for x by dividing both sides of the equation by 2:
x = 90° / 2
x = 45°
Therefore, the value of x that satisfies the equation cos(5x-10°) = sin(20-3x) is x = 45°.