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3.4 Find the general solution in each of the following {:[3.4.1,3sin theta=2cos theta],[3.4.2,2sin^(2)y-3cos y=4]:}

33.44 Find the general solution in each of the following\newline\begin{align*}\(\newline&3.4.1, &3\sin \theta=2\cos \theta,(\newline\)&3.4.2, &2\sin^{2}y-3\cos y=4\newline\end{align*}\)

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Sine and cosine of complementary anglesright chevron icon

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Q. 33.44 Find the general solution in each of the following\newline\begin{align*}\(\newline&3.4.1, &3\sin \theta=2\cos \theta,(\newline\)&3.4.2, &2\sin^{2}y-3\cos y=4\newline\end{align*}\)
  1. Identify Relationship: Identify the relationship between sin⁡\sin and cos⁡\cos in the first equation.\newlineUsing the identity sin⁡(θ)=cos⁡(90°−θ)\sin(\theta) = \cos(90° - \theta), rewrite the equation:\newline3sin⁡(θ)=2cos⁡(θ)3\sin(\theta) = 2\cos(\theta)\newline⇒3sin⁡(θ)=2sin⁡(90°−θ)\Rightarrow 3\sin(\theta) = 2\sin(90° - \theta)
  2. Solve for Theta: Solve for θ\theta by dividing both sides by sin⁡(θ)\sin(\theta), assuming sin⁡(θ)≠0\sin(\theta) \neq 0.\newlinetan⁡(θ)=23\tan(\theta) = \frac{2}{3}\newlineθ=arctan⁡(23)+kπ\theta = \arctan\left(\frac{2}{3}\right) + k\pi, where kk is any integer
  3. Address Second Equation: Address the second equation, 2sin⁡2(y)−3cos⁡(y)=42\sin^2(y) - 3\cos(y) = 4. Rewrite the equation in terms of sin⁡2(y)\sin^2(y) and cos⁡(y)\cos(y): 2sin⁡2(y)−3cos⁡(y)=42\sin^2(y) - 3\cos(y) = 4
  4. Use Pythagorean Identity: Use the Pythagorean identity sin⁡2(y)+cos⁡2(y)=1\sin^2(y) + \cos^2(y) = 1 to substitute for sin⁡2(y)\sin^2(y).\newlinesin⁡2(y)=1−cos⁡2(y)\sin^2(y) = 1 - \cos^2(y)\newlinePlug this into the equation:\newline2(1−cos⁡2(y))−3cos⁡(y)=42(1 - \cos^2(y)) - 3\cos(y) = 4\newline⇒2−2cos⁡2(y)−3cos⁡(y)=4\Rightarrow 2 - 2\cos^2(y) - 3\cos(y) = 4\newline⇒2cos⁡2(y)+3cos⁡(y)−2=0\Rightarrow 2\cos^2(y) + 3\cos(y) - 2 = 0
  5. Solve Quadratic Equation: Solve the quadratic equation for cos⁡(y)\cos(y). Using the quadratic formula, cos⁡(y)=−b±b2−4ac2a\cos(y) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} a=2a = 2, b=3b = 3, c=−2c = -2 cos⁡(y)=−3±32−4⋅2⋅(−2)2⋅2\cos(y) = \frac{-3 \pm \sqrt{3^2 - 4\cdot2\cdot(-2)}}{2\cdot2} cos⁡(y)=−3±9+164\cos(y) = \frac{-3 \pm \sqrt{9 + 16}}{4} cos⁡(y)=−3±54\cos(y) = \frac{-3 \pm 5}{4} cos⁡(y)=12\cos(y) = \frac{1}{2} or cos⁡(y)=−2\cos(y) = -2

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