- #1
fan_boy17
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Homework Statement
solve 5sinx +12cosx =6.5 between 0 and 180 degrees
Homework Equations
The Attempt at a Solution
i tried squaring both sides. (5sinx +12cosx)^2= (6.5)^2
25sinx +60sinxcosx +60sinxcosx + 144cosx^2 =42.25
Karamata said:Or, you can divide equation by [itex]\sqrt{5^2+12^2} = 13[/itex], and you will have
[itex]\frac{5}{13}\sin x + \frac{12}{13}\cos x = \frac{6.5}{13} = \frac{1}{2}[/itex]
And you know that [itex]\cos \arccos \frac{5}{13} = \frac{5}{13}[/itex] and [itex]\sin \arccos \frac{5}{13} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \frac{12}{13}[/itex]
And then, use formula for [itex]\sin(x+y)[/itex]
sorry for bad English
The range of 0-180 degrees indicates that the equation is being solved for values of x within the first two quadrants of the unit circle.
Yes, this equation can be solved algebraically by using trigonometric identities and solving for x.
Yes, the most common method for solving equations with both sine and cosine is to use the double angle identity for cosine and the half angle identity for sine.
The general process for solving this type of trigonometric equation involves isolating either sine or cosine on one side of the equation, using trigonometric identities to simplify the equation, and then using inverse trigonometric functions to solve for the variable.
Yes, there are special cases to consider when solving this equation, such as when the coefficient of sine and cosine are the same, or when the equation can be rewritten in terms of tangent.