# SOLUTION: Solve the equation 4sin(2theta)-3cos=0 for all values on the interval {{{"0°"<=theta<"360°"}}}. Express any non-integer answers to the nearest tenth of a degree.

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: Solve the equation 4sin(2theta)-3cos=0 for all values on the interval {{{"0°"<=theta<"360°"}}}. Express any non-integer answers to the nearest tenth of a degree.      Log On

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 Click here to see ALL problems on Trigonometry-basics Question 282663: Solve the equation 4sin(2theta)-3cos=0 for all values on the interval . Express any non-integer answers to the nearest tenth of a degree.Answer by Edwin McCravy(8882)   (Show Source): You can put this solution on YOUR website!``` You left off the angle for the cosine, so I can't tell whether you meant the equation to be this: or this: So I'll do it both ways: ----------------------------- If it is the first way, Use the identity Factor out Use the zero-factor principle. Set the first factor = 0: The only angles between 0° and 360° which have their cosine equaling to zero are 90° and 270°. So those are two of the solutions. Set the second factor = 0: With a calculator we find the inverse sine of is the first quadrant angle 22.02431284°. However the angle in the second quadrant whioch has 22.02431284° as its reference angle is 157.9756872°. So the four solutions on the interval , with the decimals rounded to the nearest tenth of a degree are: , , , ------------------------ If it was supposed to have been the second way: Divide both sides by Divide both sides by 4: Use the identity With a calculator we find the inverse tangent of is the first quadrant angle 36.86989765°. However the angle in the third quadrant which has 36.86989765° as its reference angle is 216.8698976°. Now since the angle is and not just , we must find all solutions for in twice the interval, that is the interval , so that will be in the interval . So we must add 360° to each of those two values, so that when we find by dividing by 2 we will have all the solutions in the interval . So we have , , , , Now solving for , we have: , , , , Or rounded to the nearest tenth of a degree: , , , . Edwin```