Question

2sinϴ=2-cosϴ then sinϴ=?

Answer 1

1, 3/5

Answer 2

We are asked to find the value(s) of $\sin\theta$ given the equation $2\sin\theta = 2 - \cos\theta$.

The given equation is $2\sin\theta = 2 - \cos\theta$. Rearrange the equation: $2\sin\theta + \cos\theta = 2$

To solve this equation, we can square both sides. First, isolate one of the terms if needed, but in this form, squaring directly might introduce $\sin\theta\cos\theta$ term. A better approach for squaring is to move one term to the other side if necessary to eliminate the constant term, or express one trig function in terms of the other.

Let's use the identity $\sin^2\theta + \cos^2\theta = 1$. We can express $\cos\theta$ in terms of $\sin\theta$ or vice versa after squaring. From the original equation, $\cos\theta = 2 - 2\sin\theta$. Substitute this into the identity $\sin^2\theta + \cos^2\theta = 1$: $\sin^2\theta + (2 - 2\sin\theta)^2 = 1$ $\sin^2\theta + (4 - 8\sin\theta + 4\sin^2\theta) = 1$ $\sin^2\theta + 4 - 8\sin\theta + 4\sin^2\theta = 1$ Combine like terms: $5\sin^2\theta - 8\sin\theta + 4 = 1$ $5\sin^2\theta - 8\sin\theta + 3 = 0$

This is a quadratic equation in terms of $\sin\theta$. Let $x = \sin\theta$. The equation becomes $5x^2 - 8x + 3 = 0$. We can solve this quadratic equation for $x$ using factorization or the quadratic formula. Using factorization: We look for two numbers that multiply to $5 \times 3 = 15$ and add up to $-8$. These numbers are $-3$ and $-5$. $5x^2 - 5x - 3x + 3 = 0$ $5x(x - 1) - 3(x - 1) = 0$ $(5x - 3)(x - 1) = 0$

This gives two possible solutions for $x = \sin\theta$:

1. $5x - 3 = 0 \implies 5x = 3 \implies x = \frac{3}{5}$ So, $\sin\theta = \frac{3}{5}$.

2. $x - 1 = 0 \implies x = 1$ So, $\sin\theta = 1$.

Since we squared the equation, we need to check if these values of $\sin\theta$ satisfy the original equation $2\sin\theta = 2 - \cos\theta$.

Case 1: $\sin\theta = 1$ If $\sin\theta = 1$, then $\theta = \frac{\pi}{2} + 2k\pi$ for some integer $k$. For these values of $\theta$, $\cos\theta = \cos(\frac{\pi}{2} + 2k\pi) = 0$. Substitute $\sin\theta = 1$ and $\cos\theta = 0$ into the original equation: $2(1) = 2 - 0$ $2 = 2$ This is true. So, $\sin\theta = 1$ is a valid solution.

Case 2: $\sin\theta = \frac{3}{5}$ If $\sin\theta = \frac{3}{5}$, then $\cos^2\theta = 1 - \sin^2\theta = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$. So, $\cos\theta = \sqrt{\frac{16}{25}} = \frac{4}{5}$ or $\cos\theta = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$. We need to check which value of $\cos\theta$ corresponds to $\sin\theta = \frac{3}{5}$ in the original equation.

Substitute $\sin\theta = \frac{3}{5}$ into the original equation $2\sin\theta = 2 - \cos\theta$: $2(\frac{3}{5}) = 2 - \cos\theta$ $\frac{6}{5} = 2 - \cos\theta$ $\cos\theta = 2 - \frac{6}{5} = \frac{10}{5} - \frac{6}{5} = \frac{4}{5}$.

So, when $\sin\theta = \frac{3}{5}$, the original equation requires $\cos\theta = \frac{4}{5}$. This pair $(\sin\theta = \frac{3}{5}, \cos\theta = \frac{4}{5})$ is consistent with $\sin^2\theta + \cos^2\theta = (\frac{3}{5})^2 + (\frac{4}{5})^2 = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$. So, $\sin\theta = \frac{3}{5}$ is a valid solution.

The possible values for $\sin\theta$ are $1$ and $\frac{3}{5}$.

Explanation of the solution: The equation $2\sin\theta = 2 - \cos\theta$ is solved by substituting $\cos\theta = 2 - 2\sin\theta$ into the identity $\sin^2\theta + \cos^2\theta = 1$. This yields a quadratic equation in $\sin\theta$: $5\sin^2\theta - 8\sin\theta + 3 = 0$. Factoring the quadratic gives $(5\sin\theta - 3)(\sin\theta - 1) = 0$, leading to $\sin\theta = \frac{3}{5}$ or $\sin\theta = 1$. Both solutions are verified by substituting back into the original equation.