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How do you find the linearization function of \displaystyle{f{{\left({x}\right)}}}={\sin{{\left({4}{x}\right)}}}+{\cos{{\left({x}\right)}}} ?https://socratic.org/questions/how-do-you-find-the-linearization-function-of-f-x-sin-4x-cos-x Steve M Jul 26, 2017 We have: \displaystyle{f{{\left({x}\right)}}}={\sin{{4}}}{x}+{\cos{{x}}} Here we will obtain an linear approximation (about \displaystyle{x}={0} ) using ... How do I solve \displaystyle{4}{\sin{{x}}}+{9}{\cos{{x}}}={0} for \displaystyle{0}{<}{x}{<}{360} ?https://socratic.org/questions/how-do-i-solve-4-sin-x-9-cos-x-0-for-0-x-360 Refer to explanation Explanation: We have that \displaystyle{4}{\sin{{x}}}+{9}{\cos{{x}}}={0}\Rightarrow{4}{\sin{{x}}}=−{9}{\cos{{x}}}\Rightarrow\frac{{\sin{{x}}}}{{\cos{{x}}}}=−\frac{{9}}{{4}}\Rightarrow{\tan{{x}}}=−\frac{{9}}{{4}} ... How do you solve \displaystyle{\sin{{2}}}{x}+{\cos{{\left(-{x}\right)}}}={0} ?https://socratic.org/questions/how-do-you-solve-sin-2x-cos-x-0-1 In the interval \displaystyle{\left[{0},{2}\pi\right]} , the solutions are \displaystyle{x}=\frac{\pi}{{2}},{7}\frac{\pi}{{6}}, and \displaystyle{11}\frac{\pi}{{6}} Explanation: It ... Why this this approach to solving a trig equation yield invalid solutions alongside valid ones?https://math.stackexchange.com/questions/1319003/why-this-this-approach-to-solving-a-trig-equation-yield-invalid-solutions-alongs First you multiplied by \sin{4x}. When this is zero, this makes the equation satisfied automatically. That's 8 extra roots... Also, because you squared, you lost the information that \sin{4x}+\cos{4x} = +1: ... Solving the trigonometric equation \sin(2x)\sec x + 2\cos x = 0 for 0 \leq x < 2\pihttps://math.stackexchange.com/questions/1962953/solving-the-trigonometric-equation-sin2x-sec-x-2-cos-x-0-for-0-leq-x 0=\sin2x\sec x+2\cos x=2(\sin x+\cos x)\implies \sin x+\cos x=0 and now observe that when \;\cos x=0\; we do not get a solution as sine and cosine do not vanish on the same points, thus for ... How do you solve this trig equation, I've tried what I know..https://math.stackexchange.com/q/586568 You have to use the formula: a\cos x + b\sin x = \sqrt{\left(a^2+b^2\right)} \cos(x - y) where y = \tan^{-1}\left(\dfrac ba\right). Using it, we get: 2\sin (x) + \cos (x) = \sqrt5\cos\left(x - \tan^{-1}(2)\right) ...More Items
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Quadratic equation { x } ^ { 2 } - 4 x - 5 = 0Trigonometry 4 \sin \theta \cos \theta = 2 \sin \thetaLinear equation y = 3x + 4Arithmetic 699 * 533Matrix \left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]Simultaneous equation \left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.Differentiation \frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }Integration \int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d xLimits \lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}Back to top
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Graph Both Sides in 2DGraph in 2DQuizTrigonometry5 problems similar to: \sin 4 x + \cos 4 x = 0
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