Question

solve each equation on the interval 0°,360°).
solving trigonometric equations using substitutio...i
share
sin left( \frac { x } { 2 }
ight) = sqrt { 2 } - sin left( \frac { x } { 2 }
ight)
u = \frac { x } { 2 }
ightarrow x = 2 u
sin ( u ) = sqrt { 2 } - sin ( u )
\frac { 2 sin ( u ) } { 2 } = \frac { sqrt { 2 } } { 2 }
sin ( u ) = \frac { sqrt { 2 } } { 2 }
u = 45 ^ { circ }
ightarrow x _ { 1 } = 2 ( 45 ) = 90 ^ { circ }
x _ { 2 } = 2 ( 135 ^ { circ } ) = 270 ^ { circ }
watch on youtube
solve ( 8 cos ( 6 x ) = 4 ) for the smallest positive solution.
give your answer accurate to at least two decimal places.

Explanation:

Step1: Simplify the equation

Divide both sides of \(8\cos(6x) = 4\) by \(8\).
\(\cos(6x)=\frac{4}{8}=\frac{1}{2}\)

Step2: Find the general solution for \(6x\)

We know that if \(\cos\theta=\frac{1}{2}\), then \(\theta = 2n\pi\pm\frac{\pi}{3}\), \(n\in\mathbb{Z}\). So for \(\theta = 6x\), we have \(6x = 2n\pi\pm\frac{\pi}{3}\).

Step3: Solve for \(x\)

Divide both sides by \(6\): \(x=\frac{2n\pi\pm\frac{\pi}{3}}{6}=\frac{n\pi}{3}\pm\frac{\pi}{18}\)

Step4: Find the smallest positive solution

When \(n = 0\):

• For \(x=\frac{n\pi}{3}+\frac{\pi}{18}\), \(x=\frac{\pi}{18}\approx 0.17\)
• For \(x=\frac{n\pi}{3}-\frac{\pi}{18}\), \(x =-\frac{\pi}{18}\) (not positive)

When \(n = 1\):

• \(x=\frac{\pi}{3}+\frac{\pi}{18}=\frac{6\pi + \pi}{18}=\frac{7\pi}{18}\approx1.22\)

Answer:

\(0.17\)