![algebra precalculus - Unclear why the highest argument value for cosine function $\cos(x+\frac{\pi}{4})$ is $\frac{\pi}{2}$ - Mathematics Stack Exchange algebra precalculus - Unclear why the highest argument value for cosine function $\cos(x+\frac{\pi}{4})$ is $\frac{\pi}{2}$ - Mathematics Stack Exchange](https://i.stack.imgur.com/TONmm.png)
algebra precalculus - Unclear why the highest argument value for cosine function $\cos(x+\frac{\pi}{4})$ is $\frac{\pi}{2}$ - Mathematics Stack Exchange
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Mathematics | Free Full-Text | A Review of Energy and Sustainability KPI-Based Monitoring and Control Methodologies on WWTPs
![đk x $\neq$ $\pi$/3 + k $\pi$ x $\neq$ $\pi$ /2 +k $\pi$ pt có nghiệm x= ± pi/3 + k2pi, x =k2pi , x=pi/2+k2pi đối chiếu điều kiện với ạ vẽ đk x $\neq$ $\pi$/3 + k $\pi$ x $\neq$ $\pi$ /2 +k $\pi$ pt có nghiệm x= ± pi/3 + k2pi, x =k2pi , x=pi/2+k2pi đối chiếu điều kiện với ạ vẽ](https://img.hoidap247.com/picture/answer/20191001/large_1569946443875.jpg)
đk x $\neq$ $\pi$/3 + k $\pi$ x $\neq$ $\pi$ /2 +k $\pi$ pt có nghiệm x= ± pi/3 + k2pi, x =k2pi , x=pi/2+k2pi đối chiếu điều kiện với ạ vẽ
Equation sin 7x + cos 2x = 2 Solution is A) x=(2kpi)/7 + 3pi/14,kI B) x=npi + pi/4 , n I C)x= 2npi + pi/2 ,n I D)none of these
![Trigonometric equations] When solving this one algebraically I get no real solutions but drawing it I see two solutions?? : r/learnmath Trigonometric equations] When solving this one algebraically I get no real solutions but drawing it I see two solutions?? : r/learnmath](https://external-preview.redd.it/UjrRWVnwLqOb2pT96wjoAMtfwjWU4C1K_ieRb7_UkIM.jpg?auto=webp&s=99eb03d5c57ab15b9663c8ae393ed97d94c3261c)