Determine the general solution of the following equation: 6 cos²x - 7 cos x − 3 = 0

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Question 1 Given sin x = 3/5, find cos x. Step 1: Start with the Pythagorean identity. sin² x + cos² x = 1 Step 2: Substitute the given value. ( 3/5 )² + cos² x = 1; 9/25 + cos² x = 1 Step 3: Isolate cos² x. cos² x = 1 − 9/25 = 16/25 Step 4: Take the square root (principal value for acute angle). cos x = ± 4/5; Assuming x is acute, cos x = 4/5. cos x = 4/5 Question 2 Given cos x = 4/5, find sin x. Step 1: Start with the Pythagorean identity. sin² x + cos² x = 1 Step 2: Substitute the given value. sin² x + ( 4/5 )² = 1; sin² x + 16/25 = 1 Step 3: Isolate sin² x. sin² x = 1 − 16/25 = 9/25 Step 4: Take the square root (principal value for acute angle). sin x = ± 3/5; Assuming x is acute, sin x = 3/5. sin x = 3/5 Question 4.1 Simplify sin(90° − θ). Step 1: In a right triangle, sin(90° − θ) is opposite over hypotenuse for angle (90° − θ). The opposite side to (90° − θ) is the adjacent side to θ. Step 2: Thus, sin(90° − θ) = adjacent to θhypotenuse = cos θ. sin(90° − θ) = cos θ Question 4.2 Simplify cos(90° − θ). Step 1: cos(90° − θ) is adjacent over hypotenuse for angle (90° − θ). The adjacent side to (90° − θ) is the opposite side to θ. Step 2: Thus, cos(90° − θ) = opposite to θhypotenuse = sin θ. cos(90° − θ) = sin θ Question 4.3 Simplify tan(90° − θ). Step 1: tan(90° − θ) = opposite to (90° − θ)adjacent to (90° − θ). Step 2: Opposite to (90° − θ) is adjacent to θ, adjacent to (90° − θ) is opposite to θ. Step 3: Thus, tan(90° − θ) = adjacent to θopposite to θ = cot θ. tan(90° − θ) = cot θ Question 4.4 Prove sin² θ + cos² θ = 1 using the right triangle. Step 1: Label the right triangle with hypotenuse r, opposite side to θ is y, adjacent side is x. sin θ = y/r, cos θ = x/r Step 2: By Pythagorean theorem in the triangle, r² = x² + y² Step 3: Divide both sides by r². (r²)/(r²) = (x²)/(r²) + (y²)/(r²); 1 = cos² θ + sin² θ; sin² θ + cos² θ = 1

Determine the general solution of the following equation: 6 cos²x - 7 cos x − 3 = 0

Step-by-step solution

Part 1· 4 steps

Start with the Pythagorean identity.

Substitute the given value.

Isolate $\cos^2 x$ .

Take the square root (principal value for acute angle).

Part 2· 4 steps

Start with the Pythagorean identity.

Substitute the given value.

Isolate $\sin^2 x$ .

Take the square root (principal value for acute angle).

Part 4.1· 2 steps

In a right triangle, $\sin(90^\circ - \theta)$ is opposite over hypotenuse for angle $(90^\circ - \theta)$ .

Thus, $\sin(90^\circ - \theta) = \frac{\text{adjacent to } \theta}{\text{hypotenuse}} = \cos \theta$ .

Part 4.2· 2 steps

$\cos(90^\circ - \theta)$ is adjacent over hypotenuse for angle $(90^\circ - \theta)$ .

Thus, $\cos(90^\circ - \theta) = \frac{\text{opposite to } \theta}{\text{hypotenuse}} = \sin \theta$ .

Part 4.3· 3 steps

$\tan(90^\circ - \theta) = \frac{\text{opposite to } (90^\circ - \theta)}{\text{adjacent to } (90^\circ - \theta)}$ .

Opposite to $(90^\circ - \theta)$ is adjacent to $\theta$ , adjacent to $(90^\circ - \theta)$ is opposite to $\theta$ .

Thus, $\tan(90^\circ - \theta) = \frac{\text{adjacent to } \theta}{\text{opposite to } \theta} = \cot \theta$ .

Part 4.4· 3 steps

Label the right triangle with hypotenuse $r$ , opposite side to $\theta$ is $y$ , adjacent side is $x$ .

By Pythagorean theorem in the triangle,

Divide both sides by $r^2$ .

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Determine the general solution of the following equation: 6 cos²x - 7 cos x − 3 = 0

Step-by-step solution

Part 1· 4 steps

Start with the Pythagorean identity.

Substitute the given value.

Isolate cos⁡2x\cos^2 xcos2x.

Take the square root (principal value for acute angle).

Part 2· 4 steps

Start with the Pythagorean identity.

Substitute the given value.

Isolate sin⁡2x\sin^2 xsin2x.

Take the square root (principal value for acute angle).

Part 4.1· 2 steps

In a right triangle, sin⁡(90∘−θ)\sin(90^\circ - \theta)sin(90∘−θ) is opposite over hypotenuse for angle (90∘−θ)(90^\circ - \theta)(90∘−θ).

Thus, sin⁡(90∘−θ)=adjacent to θhypotenuse=cos⁡θ\sin(90^\circ - \theta) = \frac{\text{adjacent to } \theta}{\text{hypotenuse}} = \cos \thetasin(90∘−θ)=hypotenuseadjacent to θ​=cosθ.

Part 4.2· 2 steps

cos⁡(90∘−θ)\cos(90^\circ - \theta)cos(90∘−θ) is adjacent over hypotenuse for angle (90∘−θ)(90^\circ - \theta)(90∘−θ).

Thus, cos⁡(90∘−θ)=opposite to θhypotenuse=sin⁡θ\cos(90^\circ - \theta) = \frac{\text{opposite to } \theta}{\text{hypotenuse}} = \sin \thetacos(90∘−θ)=hypotenuseopposite to θ​=sinθ.

Part 4.3· 3 steps

tan⁡(90∘−θ)=opposite to (90∘−θ)adjacent to (90∘−θ)\tan(90^\circ - \theta) = \frac{\text{opposite to } (90^\circ - \theta)}{\text{adjacent to } (90^\circ - \theta)}tan(90∘−θ)=adjacent to (90∘−θ)opposite to (90∘−θ)​.

Opposite to (90∘−θ)(90^\circ - \theta)(90∘−θ) is adjacent to θ\thetaθ, adjacent to (90∘−θ)(90^\circ - \theta)(90∘−θ) is opposite to θ\thetaθ.

Thus, tan⁡(90∘−θ)=adjacent to θopposite to θ=cot⁡θ\tan(90^\circ - \theta) = \frac{\text{adjacent to } \theta}{\text{opposite to } \theta} = \cot \thetatan(90∘−θ)=opposite to θadjacent to θ​=cotθ.

Part 4.4· 3 steps

Label the right triangle with hypotenuse rrr, opposite side to θ\thetaθ is yyy, adjacent side is xxx.

By Pythagorean theorem in the triangle,

Divide both sides by r2r^2r2.

Still stuck on this one?

More Mathematics Questions

Isolate $\cos^2 x$ .

Isolate $\sin^2 x$ .

In a right triangle, $\sin(90^\circ - \theta)$ is opposite over hypotenuse for angle $(90^\circ - \theta)$ .

Thus, $\sin(90^\circ - \theta) = \frac{\text{adjacent to } \theta}{\text{hypotenuse}} = \cos \theta$ .

$\cos(90^\circ - \theta)$ is adjacent over hypotenuse for angle $(90^\circ - \theta)$ .

Thus, $\cos(90^\circ - \theta) = \frac{\text{opposite to } \theta}{\text{hypotenuse}} = \sin \theta$ .

$\tan(90^\circ - \theta) = \frac{\text{opposite to } (90^\circ - \theta)}{\text{adjacent to } (90^\circ - \theta)}$ .

Opposite to $(90^\circ - \theta)$ is adjacent to $\theta$ , adjacent to $(90^\circ - \theta)$ is opposite to $\theta$ .

Thus, $\tan(90^\circ - \theta) = \frac{\text{adjacent to } \theta}{\text{opposite to } \theta} = \cot \theta$ .

Label the right triangle with hypotenuse $r$ , opposite side to $\theta$ is $y$ , adjacent side is $x$ .

Divide both sides by $r^2$ .