Find the general solution, in degrees, of the equation 2sin(3x+45°)=1. Use your general solution to find the solution of 2sin(3x+45°)=1 that is closest to 200 °.
The first thing to do when we are facing a trigonometry problem is to to draw the sine and cosine diagram on the unit circle. That takes very little time to do and is incredibly helpful to visualize the periodicity of our solutions. The equation can be written as sin(3x+45°)=1/2. We see from the diagram that between 0° and 360° there are two angles which have a sine equal to 1/2, these are 30° and 150°. Of course if we go around the unit circle by 360° as many times as we want we obtain two new solutions, therefore we have solutions: 3x1 + 45°= 30°+ 360°n 3x2 + 45° = 150° + 360°n where n is an integer number we solve and find x1= 120°n -5° x2= 120°n +35° The closest solution to 200° is the one in form x1 with n=2, so x=235°.
FS
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