given that tanx=sinx/cosx solve the following equation forx in the interveal $$ 0\degree\le x\le360\degree $$;2sin^2x-cosx=1

We are given the equation:

\[
2\sin^2x - \cos x = 1
\]

We need to solve this equation for $x$ in the interval $ 0^\circ \leq x \leq 360^\circ $.

### Step 1: Substituting for $ \sin^2 x $
We can use the Pythagorean identity:
\[
\sin^2 x = 1 - \cos^2 x
\]
Substitute $ \sin^2 x $ into the equation:

\[
2(1 - \cos^2 x) - \cos x = 1
\]

### Step 2: Simplifying the equation
Expand and simplify:

\[
2(1) - 2\cos^2 x - \cos x = 1
\]

\[
2 - 2\cos^2 x - \cos x = 1
\]

Now, subtract 1 from both sides:

\[
1 - 2\cos^2 x - \cos x = 0
\]

### Step 3: Rearranging the equation
Rearrange it as a quadratic equation in terms of $ \cos x $:

\[
2\cos^2 x + \cos x - 1 = 0
\]

### Step 4: Solving the quadratic equation
We can solve this quadratic equation using the quadratic formula. The quadratic formula is given by:

\[
\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation $ 2\cos^2 x + \cos x - 1 = 0 $, the coefficients are $a = 2$, $b = 1$, and $c = -1$. Substituting into the quadratic formula:

\[
\cos x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}
\]
\[
\cos x = \frac{-1 \pm \sqrt{1 + 8}}{4}
\]
\[
\cos x = \frac{-1 \pm \sqrt{9}}{4}
\]
\[
\cos x = \frac{-1 \pm 3}{4}
\]

This gives us two solutions for $ \cos x $:

1. $ \cos x = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} $
2. $ \cos x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 $

### Step 5: Finding the angles

#### Case 1: $ \cos x = \frac{1}{2} $
The angles where $ \cos x = \frac{1}{2} $ in the interval $ 0^\circ \leq x \leq 360^\circ $ are:
\[
x = 60^\circ, 300^\circ
\]

#### Case 2: $ \cos x = -1 $
The angle where $ \cos x = -1 $ in the interval $ 0^\circ \leq x \leq 360^\circ $ is:
\[
x = 180^\circ
\]

### Step 6: Final solution
The solutions for $ x $ are:
\[
x = 60^\circ, 180^\circ, 300^\circ
\]

Would you like more details or have any questions?

Here are 5 related questions for practice:
1. Solve $ 2\cos^2 x - \sin x = 0 $ in the interval $ 0^\circ \leq x \leq 360^\circ $.
2. Solve $ \sin 2x = \cos x $ in the same interval.
3. Find the general solution for $ \sin^2 x - 2\sin x = 0 $.
4. Solve $ 3\cos^2 x - 4\cos x + 1 = 0 $ in $ 0^\circ \leq x \leq 360^\circ $.
5. Solve $ \cos^2 x + \sin x = 1 $ in the given interval.

**Tip:** When solving trigonometric equations, remember to consider all possible angles within the given interval that satisfy the equation!

Given that tanx = sinx/cosx, solve the following equation for x in the interval 0° ≤ x ≤ 360°: 2sin²x - cosx = 1

Learn how to solve the trigonometric equation 2sin²x - cosx = 1 for x in the interval 0° ≤ x ≤ 360°. This solution involves using the Pythagorean identity and solving a quadratic equation. Suitable for high school students in Grades 10-12.

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