Problem 161 - Algebra and Trigonometry

Question

Solve each equation on the interval \([0,2 \pi)\) Do not use a calculator. $$ \sin 3 x+\sin x+\cos x=0 $$

Step-by-Step Solution

Answer

The solutions to the equation \(\sin 3x+\sin x+\cos x=0\) in the interval [0,2π) are \(x = 0, \pi, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}\)

Step 1: Trigonometric Identity manipulation

First, it’s better to convert the cosine function into a sine function so that we only have to deal with one trigonometric function. Use the relation \(\cos x = \sin \left(\frac{\pi}{2}-x\right)\) to obtain: \(\sin 3x + \sin x + \sin \left(\frac{\pi}{2} - x\right)=0\)

Step 2: Using sum-to-product formulas

Further simplify the equation using sum-to-product formulas, \(\sin a+ \sin b = 2\sin\left(\frac{a+b}{2}\right) \cos \left(\frac{a-b}{2}\right)\): \[2\sin\left(\frac{3x+x}{2}\right) \cos\left(\frac{3x-x}{2}\right) + \sin \left(\frac{\pi}{2} - x\right)= 0\] This simplifies to: \[2\sin 2x \cos x + \sin \left(\frac{\pi}{2} - x\right)= 0\]

Step 3: Using product-to-sum formulas

Convert the product of sines into a sum using the product-to-sum formula \[2 \sin a \cos b = \sin(a+b) + \sin(a-b)\] This gives: \[\sin 3x +\sin x + \sin \left(\frac{\pi}{2} - x\right)= 0\] Which simplifies to: \[\sin 3x +2\sin x= 0\]

Step 4: Solving the equation

Setting \(\sin 3x +2\sin x= 0\) to zero gives the roots: \[\sin x = 0 \;or\; \sin 3x = 0\] The roots are: \(x = 0, \pi\) for \(\sin x = 0\) and \(x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}\) for \(\sin 3x = 0\).

Step 5: Checking the interval

Lastly, it is important to check that the roots are within the given interval [0,2π). All these solutions fall into that interval, so they are all valid.