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Chapter 19: Problem 1876

The number of solution of the equation \(\sqrt{(3) \sin x+\cos x}=4\) is \(x \in[0,2 \pi]\) (a) 1 (b) 2 (c) 0 (d) 3

Short Answer

Expert verified
The number of solutions of the equation \(\sqrt{3 \sin x + \cos x} = 4\) in the interval \(x \in [0, 2 \pi]\) is (c) 0.

Step by step solution

01

Square both sides of the equation

First, we need to get rid of the square root. In order to do this, we square both sides of the equation: \((\sqrt{(3) \sin x + \cos x})^2 = (4)^2\) Simplifying the expression we get: \(3 \sin x + \cos x = 16\)
02

Rearrange the equation

Now we need to express \(\sin x\) as follows: \(\sin x = \frac{16 - \cos x}{3}\)
03

Apply trigonometric identity and solve for \(x\)

Using trigonometric identities, note that \(\sin^2x + \cos^2x = 1\). So we have: \(\frac{16^2 - 32 \cos x + \cos^2x}{3^2} + \cos^2x = 1\) Now let's solve for \(x\) within the interval \(x \in [0, 2\pi]\): \(\frac{256 - 32 \cos x + \cos^2x}{9} + \cos^2x = 1\) To simplify this expression, let's multiply both sides by \(9\): \(256 - 32 \cos x + \cos^2x + 9\cos^2x = 9\) Combine terms: \(10\cos^2x - 32\cos x + 247 = 0\)
04

Determine the number of solutions

Unfortunately, this quadratic equation cannot be factored easily, so we'll proceed by attempting to find the solutions using the quadratic formula: \(\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In this case, \(a = 10\), \(b = -32\), and \(c = 247\). Compute the discriminant: \(b^2 - 4ac = (-32)^2 - 4(10)(247) = 1024 - 9880 = -8856\) Since the discriminant is negative, the quadratic equation has no real solutions for \(\cos x\). Therefore, there are no solutions for the original equation \(\sqrt{(3) \sin x + \cos x} = 4\) within the interval \(x \in [0, 2\pi]\). The correct answer is: (c) 0

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Most popular questions from this chapter

The number of values \(x\) satisfying the equation \(\left.\left.\cot ^{-1}[\sqrt{\\{x}(x+1)\\}\right]+\cos ^{-1}\left[\sqrt{(} x^{2}+x+1\right)\right]=(\pi / 2)\) is (a) 0 (b) 1 (c) 2 (d) 3

If \(\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=(3 \pi / 2)\) then \(x^{10}+y^{10}+z^{10}+\left[3 /\left(x^{10}+y^{10}+z^{10}\right)\right]=\) (a) 0 (b) 2 (c) 4 (d) 3

\(\operatorname{cosec}\left[\tan ^{-1}\left\\{\cos \left[\cot ^{-1}(4 / \sqrt{15})\right]\right\\}\right]=\) (a) \(\sqrt{3}\) (b) \([\sqrt{(} 11) / 2]\) (c) \([\sqrt{(} 47) / 4]\) (d) \([\sqrt{(47) / 2]}\)

If \(\theta=(6 \pi / 7)\) and \(x=\tan \theta+\cot (-\theta)\) then (a) \(x>0\) (b) \(x<0\) (c) \(x=0\) (d) \(x \geq 0\)

If \(\tan (x / 2)=\operatorname{cosec} x-\sin x\) then \(\tan ^{2}(x / 2)=\) (a) \(\sqrt{5}+1\) (b) \(\sqrt{5}-1\) (c) \(\sqrt{5}-2\) (d) \(\sqrt{5}+2\)
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