Solve \( \cos \left(\frac{\pi}{2}-\theta\right)+\sin \theta+0.6=0 \) for \( 0 \leq \theta<2 \pi \)
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To solve the equation \[ \cos \left(\frac{\pi}{2}-\theta\right) + \sin \theta + 0.6 = 0, \] we can utilize the cosine co-function identity, which states that \( \cos \left(\frac{\pi}{2} - \theta\right) = \sin \theta \). Thus, we can rewrite the equation as: \[ \sin \theta + \sin \theta + 0.6 = 0. \] Combining like terms, we have: \[ 2 \sin \theta + 0.6 = 0. \] Next, we isolate \(\sin \theta\): \[ 2 \sin \theta = -0.6, \] \[ \sin \theta = -0.3. \] Now we will find \(\theta\) for \(0 \leq \theta < 2\pi\). The sine function is negative in the third and fourth quadrants. To find the angle, we first calculate the reference angle: \[ \theta_r = \arcsin(0.3). \] Using a calculator: \[ \theta_r \approx 0.30462 \quad \text{(in radians)}. \] Now, we can find the angles in the third and fourth quadrants: 1. In the third quadrant: \[ \theta = \pi + \theta_r \approx \pi + 0.30462 \approx 3.44521. \] 2. In the fourth quadrant: \[ \theta = 2\pi - \theta_r \approx 2\pi - 0.30462 \approx 5.97856. \] Thus, the solutions to the equation \( \cos \left(\frac{\pi}{2}-\theta\right)+\sin \theta+0.6=0 \) for \( 0 \leq \theta < 2\pi \) are: \[ \theta \approx 3.44521 \quad \text{and} \quad \theta \approx 5.97856. \]